Past JEE Main Entrance Paper

Hint

((b) Number of arrangement
\(
=(3 ! \times 3 ! \times 4 !) \times 3 !=(3 !)^3 4 !
\)

Hint

(c) We know, \((r+1) \cdot{ }^r P_{r-1}=(r+1) \cdot \frac{r !}{1 !}=(r+1)\) !
So, \(\left(2 \cdot{ }^1 P_0-3 \cdot{ }^2 P_1+\ldots . .51\right.[latex] terms [latex])+\)
\(
\begin{aligned}
&\quad(1 !-2 !+3 !-\ldots \text { upto } 51 \text { terms }) \\
&=[2 !-3 !+4 !-\ldots+52 !]+[1 !-2 !+3 !-\ldots+51 !] \\
&=52 !+1 !=52 !+1
\end{aligned}
\)

Hint

(d) Number of five digit numbers with 2 at \(10^{\text {th }}\) place \(=8 \times 8 \times 7 \times 6=2688\)
Q It is given that, number of five digit number with 2 at \(10^{\text {th }}\) place \(=336 \mathrm{k}\)
\(
\therefore \quad 336 k=2688 \Rightarrow k=8
\)

Hint

(d) Five digits numbers be \(1,3,5,7,9\) For selection of one digit, we have \({ }^5 C_1\) choice. And six digits can be arrange in \(\frac{6 !}{2 !}\) ways. Hence, total such numbers \(=\frac{5.6 !}{2 !}=\frac{5}{2} \cdot 6 !\)

Hint

Let the six digit number be abcdef for this number to be divisible by \(11, \mid(a+c+e)-[latex] [latex](b+d+f) \mid\) must be multiple of 11
possibility is \(a+c+e=b+d+f=12\)
Case : \(1\{a, c, e\}=\{7,5,0\}\)
\(\&\{b, d, f\}=\{9,2,1\}\)
So, number of numbers \(=2 \times 2 ! \times 3 !=24\)
Case : \(2\{a, c, e\}=\{9,2,1\}\)
\(\&\{b, d, f\}=\{7,5,0\}\)
So, number of numbers \(=3 ! \times 3 !=36\)
\(\Rightarrow[latex] total [latex]24+36\)
\(=60\)

Hint

The numbers greater than $4321$ must have $4$ or $5$ in the thousand’s place.

Case 1: Let us fix first place as 4
4 _ _ _
The second place can be filled in 2 ways \((4,5)\), third and fourth place can be filled in 6 ways.
Number of such four-digit numbers \(=2 \times 6 \times 6=72\)
Case 2: Let us fix first place as 4 and second place as 3
43 _ _
The third place can be filled in 3 ways \((3,4,5)\) and the fourth place can be filled in 6 ways.
Number of such four-digit numbers \(=3 \times 6=18\)
Case 3: Let us fix first place as 4 , second place as 3 and third place as 2 432 _
The fourth place can be filled in 4 ways \((2,3,4,5)\)
Number of such four-digit numbers \(=4\)
Case 4: Let us fix first place as 5
5 _ _ _
Number of such four-digit numbers \(=6 \times 6 \times 6=216\)
So, total number \(=72+18+4+216=310\)

Hint

\(
n=10
\)
Select any three distinct numbers by \({ }^{10} \mathrm{C}_3\), then assign the biggest number to \(n_1\), the second biggest to \(n_2\) and smallest to \(n_3\)
\(\therefore\) required number of ways in which the balls can be chosen \(={ }^{10} \mathrm{C}_3\)
\(
\begin{aligned}
&=\frac{10 !}{3 ! 7 !} \\
&=\frac{10 \times 9 \times 8}{3 \times 2} \\
&=120
\end{aligned}
\)

Hint

(a) The thousands place can only be filled with 2,3 or 4 , since the number is greater than 2000 .
For the remaining 3 places, we have pick out digits such that the resultant number is divisible by 3 .
It the sum of digits of the number is divisible by 3 , then the number itself is divisible by 3 .
Case 1: If we take 2 at thousands place.
The remaining digits can be filled as:
0,1 and 3 as \(2+1+0+3=6\) is divisible by 3 .
0,3 and 4 as \(2+3+0+4=9\) is divisible by 3 .
In both the above combinations the remaining three digits can be arranged in 3! ways.
\(\therefore\) Total number of numbers in this case \(=2 \times 3 !=12\).
Case 2: If we take 3 at thousands place. The remaining digits can be filled as:
0,1 and 2 as \(3+1+0+2=6\) is divisible by 3 .
0,2 and 4 as \(3+2+0+4=9\) is divisible by 3 .
In both the above combinations, the remaining three digits can be arranged in 3 ! ways. Total number of numbers in this case \(=2 \times 3 !=12\).
Case 3: If we take 4 at thousands place.
The remaining digits can be filled as:
0,2 and 3 as \(4+2+0+3=9\) is divisible by 3 .
In the above combination, the remaining three digits can be arranged in 3! ways.
\(\therefore\) Total number of numbers in this case \(=3 !=6\).
\(\therefore\) Total number of numbers between 2000 and 5000 divisible by 3 are \(12+12+6=30\).

Hint

The number of ways of arranging 5 boys and 3 girls, i.e. 8 people on a round table would be 7 !

We subtract the number of ways of arranging those people when \(B_1\) and \(G_1\) are always together to get the required answer.

When \(B_1\) and \(G_1\) are together, we get 4 boys \(+2\) girls \(+1\left(B_1+G_1\right)\) i.e. 7 people and since \(B_1+G_1\) can be permuted in 2 ways, these can be arranged in 6 ! \(x 2\) ways.
Subtracting, we have \(7 !-6 ! \times 2=6 !(7-2)\)
\(=5 \times 6\) ! ways in total.

Alternate:
(a) 4 boys and 2 girls in circle
\(
\Rightarrow 5 ! \times \frac{6 !}{4 ! 2 !} \times 2 ! \Rightarrow 5 \times 6 !
\)

Hint

Position of SMALL in a dictionary:
Let us fix A in the first position.
Then, \(\mathrm{A}\)
At the 4 places we want to arrange L,L,M and \(S\) which can be done in \(\frac{4 !}{2 !}=12\) ways. (since 2 Ls are there)
Let us fix L in the first position.
Then, \(\mathrm{L}\)
At the 4 places we want to arrange L,A,M and \(S\) which can be done in \(4 !=24\) ways.
Let us fix \(M\) in the first position.
Then, \(\mathrm{M}\)
At the 4 places we want to arrange L,L,A and \(S\) which can be done in \(\frac{4 !}{2 !}=12\) ways. (since 2 Ls are there)
Let us fix \(\mathrm{S}\) in the first position and \(\mathrm{A}\) in second position.
Then, SA
At the 3 places we want to arrange \(L, L\) and \(M\) which can be done in \(\frac{3 !}{2 !}=3\) ways.
Let us fix \(\mathrm{S}\) in the first position and \(\mathrm{L}\) in second position.
Then, S L
At the 3 places we want to arrange L,A and \(M\) which can be done in \(3 !=6\) ways.
Let us fix \(\mathrm{M}\) in second position.
Then, S M
There arranging in alphabetical order, the next word is
SMALL
Position \(=12+24+12+3+6+1=58^{\text {th }}\)

Hint

(b) M, EEE, D. I, T, RR, AA, NN \(\mathrm{R}–\mathrm{E}\)
Two empty places can be filled with identical letters[EE, AA, NN] \(\Rightarrow 3\) ways
Two empty places, can be filled with distinct letters[M, E, D, I, T, R, A, N] \(\Rightarrow{ }^8 \mathrm{P}_2\)
\(\therefore \quad\) Number of words \(3+{ }^8 \mathrm{P}_2=59\)

Alternate:

First letter is R and Fourth letter is \(\mathrm{E}\).
So now we have 11 letters left with \(2 \mathrm{E}, 2 \mathrm{~A}, 1 \mathrm{M}, 1 \mathrm{D}, 1 \mathrm{I}, 1 \mathrm{~T}, 1 \mathrm{R}\) and \(2 \mathrm{~N}\) And with this we have to form a two-letter word.
There will be 2 cases when both these two letters of the middle will be the same. And this will be when they both are either EE, AA or NN.
Now when middle two letters are not same. So we have to form a 2letter word with 8letters without repetition .
Number of such words will be \(8 \times 7=56\)
Hence, the total number of such four-letter words with \(\mathrm{R}\) as the first letter and \(\mathrm{E}\) as the last letter will be \(56+3=59\).

Hint

(b) Total number of integral points inside the square \(\mathrm{OABC}=40 \times[latex] [latex]40=1600\)
No. of integral points on AC
\(
\begin{aligned}
=& \text { No. of integral points on } \mathrm{OB} \\
=& 40[\text { namely }(1,1),(2,2) \ldots(40,40)] \\
& \quad \therefore \quad \text { No. of integral points inside the } 0 \mathrm{OAC} \\
=& \frac{1600-40}{2}=780
\end{aligned}
\)

Hint

(d) Four digits number can be arranged in \(3 \times 4\) ! ways. Five digits number can be arranged in 5! ways. Number of integers \(=3 \times 4 !+5 !=192\).

Hint

(d) Number of ways of selecting a man and a woman for a team from 15 men and 15 women \(=15 \times 15=(15)^2\)
Number of ways of selecting a man and a woman for next team out of the remaining 14 men and 14 women.
\(
=14 \times 14=(14)^2
\)
Similarly for other teams
Hence required number of ways
\(
=(15)^2+(14)^2+\ldots .+(1)^2=\frac{15 \times 16 \times 31}{6}=1240
\)

Hint

(b) With 3 at unit place, total possible four digit number (without repetition) will be \(3 !=6\)
With 4 at unit place, total possible four digit numbers will be \(3 !=6\)
With 5 at unit place, total possible four digit numbers will be \(3 !=6\)
With 6 at unit place, total possible four digit numbers will be \(3 !=6\)
Sum of unit digits of all possible numbers
\(
\begin{aligned}
&=6 \times 3+6 \times 4+6 \times 5+6 \times 6 \\
&=6[3+4+5+6]=6[18]=108
\end{aligned}
\)

Hint

(c) Card numbered 1 is always placed in envelope numbered 2 , we can consider two cases :
Case I: Card numbered 2 is placed in envelope numbered 1.
Then it is dearrangement of 4 objects, which can be done in \(4 !\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}\right)=9\) ways
Case II: Card numbered 2 is not placed in envelope numbered 1.
Then it is dearrangement of 5 objects, which can be done in \(5 !\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}\right)=44\) ways
\(\therefore \quad\) Total ways \(=44+9=53\)

Alternate:

(Number of derrangements of \(6=6 !\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\frac{1}{6 !}\right)\)
\(
=360-120+30-6+1
\)
\(=265\)
Out of these derrangements, there are five ways in which card numbered 1 is going wrong.
So, when it is going in envelope numbered 2 is,
\(
=\frac{265}{5}=53 \text { ways. }
\)

Hint

\(p=5 !\) (as every combination of these numbers will be greater than 20000 , as the least number that can be formed is 23579)
Now, \(q=4 !+4 !+4 !=3.4 !\)
As for the number to be greater than 30000 and less than 90000, First digit of the numbers should be 3 or 5 or 7 and remaining digits can be arranged in any order,
\(3_{—-}=4\) !
\(5_{—-}=4\) !
\(7_{—-}=4\) !
Hence Total number of ways \(=3.4 !\) )
Now, \(\frac{\mathrm{p}}{\mathrm{q}}=\frac{5 !}{3.4 !}=\frac{5}{3}\)

Hint

(c) We have to form 7 digit numbers, using the digits 1,2 and 3 only, such that the sum of the digits in a number \(=10\).
This can be done by taking \(2,2,2,1,1,1\), 1, or by taking \(2,3,1,1,1,1,1\).
\(
\therefore \text { Number of ways }=\frac{7 !}{3 ! 4 !}+\frac{7 !}{5 !}=77
\)

Alternate:

Case-1:
\(1,1,1,1,1,2,3\)
Number of ways \(=\frac{7 !}{5 !}=42\) [Since, 1 repeat five times \(]\)
Case- 2 :
\(1,1,1,1,2,2,2\)
Number of ways \(=\frac{7 !}{4 ! \times 3 !}=35\) [Since, 4 repeat four times and 2 repeat three times ]
Total number of ways \(=42+35=77\)

Hint

The given word is \(\mathrm{COCHIN}\)
So, we have two times \(C\) appearing in the word.
Fixing position these \(C\) we have number of letters for permutation \(=4\)
Therefore, the number of words, having two letter of \(C\) in first and second
places \(=4 !\)
Similarly,
The number of words starting with letter \(\mathrm{CH}=4\) !
The number of words starting with letter \(C I=4\) !
The number of words starting with letter \(C N=4\) !
Therefore,
The total number of words that appear before the word ‘COCHIN’
\(
\begin{aligned}
&=4 !+4 !+4 !+4 ! \\
&=(4 \times 3 \times 2 \times 1)+(4 \times 3 \times 2 \times 1)+(4 \times 3 \times 2 \times 1)+(4 \times 3 \times 2 \times 1) \\
&=24+24+24+24 \\
&=96
\end{aligned}
\)

Alternate:

(c) The letter of word \(C O C H I N\) in alphabetic order are \(C, C, H, I, N\), 0.
Fixing first and second letter as \(C, C\), rest 4 can be arranged in \(4 !\) ways. Similarly the words starting with each of \(C H, C I, C N\) are 4 ! Then fixing first two letters as \(C O\) and next four places when filled in alphabetic order with remaining 4 letters give the word COCHIN.
\(\therefore \quad\) Numbers of words coming before COCHIN \(=4 \times 4 !=4 \times 24=96\)

Hint

Step 1: Find the number of ways of choosing exponents of \(r\).
The LCM of \(p, q\) is \(r^2 t^4 s^2\), it means \(r^2, t^4, s^2\) must be among the prime factors of \(p, q\).
Now, if \(r^2\) is a factor of \(p\), then \(q\) has \(r^x\), where \(x=(0,1)\)
So, the number of ways of choosing exponents of \(r\) will be 2 .
Similarly, if \(r^2\) is a factor of \(q\), then \(p\) has \(r^x\), where \(x=(0,1)\)
So, the number of ways of choosing exponents of \(r\) will again be 2.
But if \(r^2\) is a factor of both \(p, q\), then the number of ways of choosing exponents of \(r\) will be 1 .
So, the exponents of \(r\) can be chosen in \(2+2+1=5\) ways.

Step 2: Find the number of ways of choosing exponents of \(t\).
If \(t^4\) is a factor of \(p\), then \(q\) has \(t^x\), where \(x=(0,1,2,3)\)
So, the number of ways of choosing exponents of \(t\) will be 4 .
Similarly, if \(t^4\) is a factor of \(q\), then \(p\) has \(t^x\), where \(x=(0,1,2,3)\)
So, the number of ways of choosing exponents of \(t\) will again be 4 .
But if \(t^4\) is a factor of both \(p, q\), then the number of ways of choosing exponents of \(t\) will be 1 .
So, the exponents of \(t\) can be chosen in \(4+4+1=9\) ways.

Step 3: Find the number of ways of choosing exponents of \(s\).
If \(s^2\) is a factor of \(p\), then \(q\) has \(s^x\), where \(x=(0,1)\)
So, the number of ways of choosing exponents of \(s\) will be 2 .
Similarly, if \(s^2\) is a factor of \(q\), then \(p\) has \(s^x\), where \(x=(0,1)\)
So, the number of ways of choosing exponents of \(s\) will again be 2 .
But if \(s^2\) is a factor of both \(p, q\), then the number of ways of choosing exponents of \(s\) will be 1 .
So, the exponents of \(s\) can be chosen in \(2+2+1=5\) ways.

Step 4: Find the number of ordered pair \((p, q)\).
The number of ordered pair \((p, q)\) is \(5 \times 9 \times 5=225\)

Hint

(a) Total number of ways of arranging the letters of the word BANANA is \(\frac{6 !}{2 ! 3 !}=60\). Number of words in which 2 N’s come together is \(\frac{5 !}{3 !}=20\)
\(\therefore\) the required number \(=60-20=40\)

Hint

(c) \(X-X-X-X-X\). The four digits \(3,3,5,5\) can be arranged at (-) places in \(\frac{4 !}{2 ! 2 !}=6\) ways
The five digits \(2,2,8,8,8\) can be arranged at (X) places in \(\frac{5 !}{2 ! 3 !}=10\) ways
\(\therefore\) Total number of arrangements \(=6 \times 10=60\) ways

Hint

(a) We know that a number is divisible by 3 if the sum of its digits is divisibly by 3.
Now out of \(0,1,2,3,4,5\) if we take \(1,2,3,4,5\) or \(0,1,2,4,5\) then the 5 digit numbers will be divisible by 3 .
Case I: Number of 5-digit numbers formed using the digits 1, 2, 3, 4, \(5=\) \(5 !=120\)
Case II : Taking \(0,1,2,4,5\) if we make 5 digit number then 1st place can be filled in 4 ways ( 0 can not come at I place)
2nd place can be filled in 4 ways
3 rd place can be filled in 3 ways
4th place can be filled in 2 ways
5 th place can be filled in 1 way
\(\therefore \quad\) Total numbers \(=4 \times 4 !=96\)
Thus total numbers divisible by 3 are \(=120+96=216\)

Hint

(a) Total number of words that can be formed using 5 letters out of 10 given different letters
\(
\begin{aligned}
&=10 \times 10 \times 10 \times 10 \times 10 \text { (as letters can repeat) } \\
&=1,00,000
\end{aligned}
\)
Number of words that can be formed using 5 different letters out of 10 different letters
\(={ }^{10} P_5\) (none can repeat)
\(
=\frac{10 !}{5 !}=30,240
\)
\(\therefore \quad\) Number of words in which at least one letter is repeated
\(=\) total words-words with none of the letters repeated
\(=1,00,000-30,240 \quad=69,760\)

Hint

maximum number of hats used of same are 2 .They can not be 3 otherwise atleast 2 hats of same colour are consecutive
Now,Let hats used are R, R, G, G, B
(which can be selected in 3 ways .it can be RGGBB or RRGBB also )
Now numbers of ways of distributing blue hat (single one) in 5 Person equal to 5 let blue hat goes to person \(A\).
Now either position \(B\) and \(D\) are filled by green hats and \(C\) and \(E\) are filled by reds hats or \(\mathrm{B}\) and \(\mathrm{D}\) are filled by Red hats and \(\mathrm{C}\) and \(\mathrm{E}\) are filled by green hats \(\Rightarrow 2\) ways are possible
Hence total number of ways \(=3 \times 5 \times 2=30\) ways

Hint

A number is divisible by 4 if last two digits of that number is divisible by 4 The 2 digits numbers which are divisible by 4 from the given set \(\{1,2,3,4,5\}\) are \(12,24,32,44\) or 52
So, if a number ends with \(12,24,32,44\) or 52 , then it is divisible by 4.
Also each of the first 3 digits can be any of {1,2,3,4,5}
Therefore 5 options for each of the first 3 digits and total 5 options for the last 2 digits.
First 3 digits can be arranged in \(5^3\) ways
So, the total number of 5-digit numbers that are divisible by 4 without repetition is
\(
5^3 \times 5=625
\)

Hint

(7) \(\because n_1, n_2, n_3, n_4\) and \(n_5\) are positive integers such that \(n_1<n_2<n_3<\) \(n_4<n_5\)
Then for \(n_1+n_2+n_3+n_4+n_5=20\)
If \(n_1, n_2, n_3, n_4\) take minimum values \(1,2,3,4\) respectively then \(n_5\) will be maximum 10 .
\(\therefore\) Corresponding to \(n_5=10\), there is only one solution \(n_1=1, n_2=2, n_3=3, n_4=4\).
Corresponding to \(n_5=9\), we can have, only solution \(n_1=1, n_2=2, n_3=3, n_4=5\) i.e., one solution
Corresponding to \(n_5=8\), we can have, only solution
\(
n_1=1, n_2=2, n_3=3, n_4=6
\)
or \(n_1=1, n_2=2, n_3=4, n_4=5\)
i.e., 2 solution
For \(n_5=7\), we can have
\(
n_1=1, n_2=2, n_3=4, n_4=6
\)
or \(n_1=1, n_2=3, n_3=4, n_4=5\)
i.e. 2 solutions
For \(n_5=6\), we can have
\(
n_1=2, n_2=3, n_3=4, n_4=5
\)
i.e., one solution
Thus there can be 7 solutions.

Hint

To find rank of MOTHER
Total words starting from \(E=5 !=120\)
Total words starting from \(\mathrm{H}=5 !=120\)
Number of words starting from \(\mathrm{ME}=4 !=24\)
Number of words starting from \(\mathrm{MH}=4 !=24\)
Number of words starting from \(\mathrm{MOE}=3 !=6\)
Number of words starting from \(\mathrm{MOH}=3 !=6\)
Number of words starting from \(\mathrm{MOR}=3 !=6\)
Number of words starting from MOTE \(=2 !=2\)
Next word \(=\) MOTHER
\(\operatorname{Rank}=120+120+24+24+6+6+6+2+1\) \(=309\)

Hint

(1080) Groups can be possible in only \(2,2,1,1\) way. Number of ways of dividing persons in group
\(
=\frac{6 !}{(2 !)^2(1 !)^2(2 !)^2}
\)
Number of ways after arranging rooms \(=\frac{6 !}{(2 !)^4} \cdot 4 !=1080\)

Hint

We know that number of dearrangements of \(n\) objects
\(
=n !\left[1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\ldots . \frac{1}{n !}\right]
\)
\(\therefore \quad\) No. of ways of putting all the 4 balls into boxes of different colour
\(
\begin{aligned}
&=4 !\left[1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}\right]=4 !\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{24}\right) \\
&=24\left(\frac{12-4+1}{24}\right)=9
\end{aligned}
\)

Hint

Number of students who gave wrong answers to exactly one question \(=a_1-a_2, \quad\) Two questions \(=a_2-a_3\) Three questions \(=a_3-a_4, k-1\) question \(=a_{k-1}-a_k, k_{\text {question }}=a_{\mathrm{k}}\)
\(\therefore\) Total number of wrong answers
\(
\begin{aligned}
&=1\left(a_1-a_2\right)+2\left(a_2-a_3\right)+3\left(a_3-a_4\right)+\ldots .(k-1) \\
&\left(a_{k-1}-a_k\right)+k a_1 \\
&=a_1+a_2+a_3+\ldots . a_k
\end{aligned}
\)

Hint

(b) Distinct \(n\) digit numbers which can be formed using digits 2,5 and 7 are \(3^n\).
We have to find \(n\) so that \(3^n \geq 900 \Rightarrow 3^{n-2} \geq 100\)
\(\Rightarrow n-2 \geq 5 \Rightarrow n \geq 7\). So the least value of \(n\) is 7

Hint

(c) Given 6 boys \(M_1, M_2, M_3, M_4, M_5, M_6\) and 5 girls \(G_1, G_2, G_3, G_4\), \(G_5\)
(i) \(\alpha_1 \rightarrow\) Total number of ways of selecting 3 boys and 2 girls from 6 boys and 5 girls
i.e., \({ }^6 \mathrm{C}_3 \times{ }^5 \mathrm{C}_2=20 \times 10=200\)
\(\therefore \quad \alpha_1=200\)
(ii) \(a_2 \rightarrow\) Total number of ways selecting at least 2 member and having equal number of boys and girls
\(
\begin{aligned}
&\text { i.e., } \begin{aligned}
{ }^6 \mathrm{C}_1^5 \mathrm{C}_1+{ }^6 \mathrm{C}_2^5 \mathrm{C}_2+{ }^6 \mathrm{C}_3^5 \mathrm{C}_3+{ }^6 \mathrm{C}_4^5 \mathrm{C}_4+{ }^6 \mathrm{C}_5{ }^5 \mathrm{C}_5 \\
&=30+150+200+75+6=461
\end{aligned} \\
&\Rightarrow \alpha_2=461
\end{aligned}
\)
(iii) \(\alpha_3 \rightarrow\) Total number of ways of selecting 5 members in which at least 2 of them girls
\(
\begin{aligned}
&\text { i.e., }{ }^5 \mathrm{C}_2{ }^6 \mathrm{C}_3+{ }^5 \mathrm{C}_3{ }^6 \mathrm{C}_2+{ }^5 \mathrm{C}_4{ }^6 \mathrm{C}_1+{ }^5 \mathrm{C}_5{ }^6 \mathrm{C}_0 \\
&=200+150+30+1=381 \\
&\Rightarrow a_3=381
\end{aligned}
\)
(iv) \(\alpha_4 \rightarrow\) Total number of ways for selecting 4 members in which at least two girls such that \(M_1\) and \(G_1\) are not included together.
\(G_1\) is included \(\rightarrow{ }^4 \mathrm{C}_1 \cdot{ }^5 \mathrm{C}_2+{ }^4 \mathrm{C}_2 \cdot{ }^5 \mathrm{C}_1+{ }^4 \mathrm{C}_3\)
\(
=40+30+4=74
\)
\(M_1\) is included \(\rightarrow{ }^4 \mathrm{C}_2 \cdot{ }^5 \mathrm{C}_1+{ }^4 \mathrm{C}_3=30+4=34\)
\(G_1\) and \(M_1\) both are not included
\(
\begin{gathered}
1+20+60=81 \quad{ }^4 \mathrm{C}_4+{ }^4 \mathrm{C}_3 \cdot{ }^5 \mathrm{C}_1+{ }^4 \mathrm{C}_2 \cdot{ }^5 \mathrm{C}_2 \\
\therefore \quad \text { Total number }=74+34+81=189 \\
\alpha_4=189
\end{gathered}
\)
Now, \(P \rightarrow 4 ; Q \rightarrow 6 ; R \rightarrow 5 ; S \rightarrow 2\)

Hint

(A) For the permutations containing the word ENDEA we consider ‘ENDEA’ as single letter. Then we have total ENDEA, N, O, E, L i.e. 5 letters which can be arranged in 5! ways.
\(
\therefore(\mathrm{A}) \rightarrow(\mathrm{p})
\)
(B) If \(\mathrm{E}\) occupies the first and last position, the middle 7 positions can be filled by \(\mathrm{N}, \mathrm{D}, \mathrm{E}, \mathrm{A}, \mathrm{N}, \mathrm{O}, \mathrm{L}\). in \(\frac{7 !}{2 !}=7 \times 6 \times 5 \times 4 \times 3=21 \times 120=21 \times 5 !\) ways. \(\therefore(\mathrm{B}) \rightarrow(\mathrm{s})\)
(C) If none of the letters D, L, N occur in the last five positions then we should arrange D, L, N, N at first four positions and rest five i.e. \(\mathrm{E}, \mathrm{E}, \mathrm{E}, \mathrm{A}, \mathrm{O}\) at last five positions. This can be done in
\(
\begin{aligned}
&\frac{4 !}{2 !} \times \frac{5 !}{3 !}=4 \times 3 \times \frac{5 !}{3 \times 2}=2 \times 5 \text { ! ways } \\
&\therefore \quad(C) \rightarrow \text { (q) }
\end{aligned}
\)
(D) As per question A, E, E, E, O can be arranged at 1st, 3rd, 5th, 7th and 9th positions and rest \(\mathrm{D}, \mathrm{L}, \mathrm{N}, \mathrm{N}\) at rest 4 positions. This can be done in
\(
\frac{5 !}{3 !} \times \frac{4 !}{2 !} \text { ways }=2 \times 5 \text { ! ways } \quad \therefore \quad \text { (D) } \rightarrow \text { (q) }
\)

Hint

Given: Runs scored in \(k^{\text {th }}\) match \(=k \cdot 2^{n+1-k}, 1 \leq k \leq n\) and runs scored in \(\mathrm{n}\) matches \(=\frac{n+1}{4}\left(2^{n+1}-n-2\right)\)
\(
\begin{array}{cl}
\therefore & \sum_{k=1}^n k \cdot 2^{n+1-k}=\frac{n+1}{4}\left(2^{n+1}-n-2\right) \\
\Rightarrow & 2^{\mathrm{n}+1}\left[\sum_{k=1}^n \frac{k}{2^k}\right]=\frac{n+1}{4}\left(2^{n+1}-n-2\right) \\
\Rightarrow & 2^{n+1}\left[\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\ldots+\frac{n}{2^n}\right] \\
& =\frac{n+1}{4}\left(2^{n+1}-n-2\right) \dots(i)
\end{array}
\)
Let \(S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\ldots .+\frac{n}{2^n} \dots(ii)\) 
\(
\therefore \quad \frac{1}{2} S=\frac{1}{2^2}+\frac{2}{2^3}+\ldots .+\frac{n-1}{2^n}+\frac{n}{2^{n+1}} \dots(iii)
\)
On subtracting eq. (iii), from (ii), we get i.e., \(\frac{1}{2} S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3} .+\ldots+\frac{1}{2^n}-\frac{n}{2^{n+1}}\)
\(
\begin{aligned}
&\Rightarrow \frac{1}{2} S=\frac{\frac{1}{2}\left(1-\frac{1}{2^n}\right)}{1-\frac{1}{2}}-\frac{n}{2^{n+1}} \\
&\Rightarrow S=2\left[1-\frac{1}{2^n}-\frac{n}{2^{n+1}}\right] \dots(iv)
\end{aligned}
\)
From equation (i) and (iv),
\(
\begin{aligned}
2.2^{\mathrm{n}+1} & {\left[1-\frac{1}{2^n}-\frac{n}{2^{n+1}}\right]=\frac{n+1}{4}\left[2^{n+1}-n-2\right] } \\
& \Rightarrow 2 .\left[2^{n+1}-2-n\right]=\frac{n+1}{4}\left[2^{n+1}-2-n\right] \\
& \Rightarrow \frac{n+1}{4}=2 \Rightarrow n=7
\end{aligned}
\)

Hint

Let there be \(n\) sets of different objects each set containing \(n\) identical objects [eg (1, 1, \(1 \ldots 1\) ( \(n\) times) ), \((2,2,2 \ldots, 2\) \((n\) times \()) \ldots(n, n, n \ldots n(n\) times \())]\)
Then the number of ways in which these \(n \times n=n^2\) objects can be arranged
\(
\text { in a row }=\frac{\left(n^2\right) !}{n ! n ! \ldots n !}=\frac{\left(n^2\right) !}{(n !)^n}
\)
But these number of ways should be a natural number.
Hence \(\frac{\left(n^2\right) !}{(n !)^n}\) is an integer. \(\left(n \in I^{+}\right)\)

Hint

Since, \(\mathrm{m}\) men can be seated in \(\mathrm{m}\) ! ways creating \((m+1)\) places for ladies to sit.
\(\therefore \quad n\) ladies out of \((m+1)\) places \((\) as \(n<m)\) can be seated in \({ }^{m+1} P_n\) ways
\(
\begin{aligned}
&\therefore \quad \text { Total ways }=m ! \times{ }^{m+1} P_n \\
&\quad=m ! \times \frac{(m+1) !}{(m+1-n) !}=\frac{(m+1) ! m !}{(m-n+1) !}
\end{aligned}
\)

Hint

Given: There are three sections containing 4,5 and 6 questions respectively.
Now, from each question, 3 questions should be answered.
Therefore total number of ways, for the selection of question can be done
\(
(1,1,3)+(1,2,2)+(1,3,1)+(2,1,2)+(2,2,1)+(3,1,1)
\)
\(\therefore\) Total number of selection of 5 questions
\(
\begin{gathered}
=3 \times{ }^5 C_1 \times{ }^5 C_1 \times{ }^5 C_3+3 \times{ }^5 C_1 \times{ }^5 C_2 \times{ }^5 C_2 \\
=3 \times 5 \times 5 \times 10+3 \times 5 \times 10 \times 10=750+1500=2250
\end{gathered}
\)

Hint

It is given that, there are metro stations located in a city along a circular path. Each pair of the nearest station is connected by a blue line whereas all remaining are connected by a red line.
Also the number of red lines is equal to 99 times the number of blue lines Therefore we have,
Number of blue lines \(=\) Number of sides \(=n\)
Number of red lines \(=\) number of diagonals \(={ }^n C_2-n\)
\(
\begin{aligned}
&{ }^n C_2-n=99 n \Rightarrow \frac{n(n-1)}{2}-n=99 n \\
&{ }^{n-1}-1=99 \Rightarrow n=201
\end{aligned}
\)

Hint

Step 1: Expressing the given data:
Given that \(a, b\) and \(c\) are the greatest values of \({ }^{19} C_p,{ }^{20} C_q,{ }^{21} C{ }_r\).
We know that
\({ }^n C_r\) is maximum when \(r= \begin{cases}\frac{n}{2} & , n \text { is even } \\ \frac{n-1}{2} \text { or } \frac{n+1}{2} & , n \text { is odd }\end{cases}\)
Therefore, \(\max \left({ }^{19} C_p\right)=\left({ }^{19} C_9\right)=a\)
\(
\begin{aligned}
&\max \left({ }^{20} C_q\right)={ }^{20} C_{10}=b \\
&\max \left({ }^{21} C_r\right)={ }^{21} C_{11}=c
\end{aligned}
\)
Step 2: Find the relation :
\(
\begin{aligned}
&\therefore \frac{a}{{ }^{19} C_9}=\frac{b}{\frac{20}{10} \times{ }^{19} C_9}=\frac{c}{\frac{21}{11} \times \frac{20}{10} \times{ }^{19} C_9} \\
&\Rightarrow \quad \frac{a}{1}=\frac{b}{2}=\frac{c}{\frac{42}{11}} \\
&\Rightarrow \frac{a}{11}=\frac{b}{22}=\frac{c}{42}
\end{aligned}
\)

Hint

Using \({ }^{36} \mathrm{C}_{r+1}=\frac{36}{r+1} \times{ }^{35} \mathrm{C}_r\), we get
\(
\begin{aligned}
&\frac{36}{r+1} \times{ }^{35} C_r \times\left(k^2-3\right)={ }^{35} C_r \times 6 \\
&\Rightarrow k^2-3=\frac{r+1}{6} \\
&\Rightarrow k^2=\frac{r+1}{6}+3 \\
&k \in I
\end{aligned}
\)
\(r \rightarrow\) Non-negative integer \(0 \leq \mathrm{r} \leq 35\)
\(
\begin{aligned}
&r=5 \Rightarrow k=\pm 2 \\
&r=35 \Rightarrow k=\pm 3
\end{aligned}
\)
\(\therefore\) No. of ordered pairs \((r, k)=4\)
Therefore, the required ordered pairs are \((5,-2),(5,2),(35,-3)\) and \((35,3)\). Therefore, The total number of required ordered pairs \((r, k)\) is 4.

Hint

Given that there are 31 Objects in which 10 are identical and 21 are distinct
The number of ways of choosing 0 identical and 10 distinct \(=1 \times{ }^{21} \mathrm{C}_{10}\)
The number of ways of choosing 1 identical and 9 distinct \(=1 \times{ }^{21} \mathrm{C}_9\)
The number of ways of choosing 2 identical and 8 distinct \(=1 \times{ }^{21} \mathrm{C}_8\)
The number of ways of choosing 3 identical and 7 distinet \(=1 \times{ }^{21} C_7\)
The number of ways of choosing 4 identical and 6 distinct \(=1 \times{ }^{21} \mathrm{C}_6\)
The number of ways of choosing 5 identical and 5 distinct \(=1 \times{ }^{21} \mathrm{C}_5\)
The number of ways of choosing 6 identical and 4 distinct \(=1 \times{ }^{21} \mathrm{C}_4\)
The number of ways of choosing 7 identical and 3 distinct \(=1 \times{ }^{21} C_3\)
The number of ways of choosing 8 identical and 2 distinct \(=1 \times{ }^{21} C_2\)
The number of ways of choosing 9 identical and 1 distinct \(=1 \times{ }^{21} C_1\)
The number of ways of choosing 10 identical and 0 distinct= \(1 \times{ }^{21} \mathrm{C}_0\)
So, total number of ways in which we can choose 10 objects is
\(
\begin{aligned}
&{ }^{21} \mathrm{C}_{10}+{ }^{21} \mathrm{C}_9+{ }^{21} \mathrm{C}_8+\cdots+{ }^{21} \mathrm{C}_0=\mathrm{x} \text { (Let) } \dots(i)\\
&\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}} \\
&{ }^{21} \mathrm{C}_{11}+{ }^{21} \mathrm{C}_{12}+{ }^{21} \mathrm{C}_{13}+\cdots+{ }^{21} \mathrm{C}_{21}=\mathrm{x} \dots(ii)
\end{aligned}
\)
On adding both Equation (i) and (ii), we get
\(
\begin{aligned}
&2 x={ }^{21} C_0+{ }^{21} C_1+{ }^{21} C_2+\cdots+{ }^{21} C_{21} \\
&\therefore 2 x=2^{21} \Rightarrow x=2^{20}
\end{aligned}
\)

Hint

It is given that, there are 20 pillars of the same height have been erected along the boundary of a circular stadium.
Now, the top of each pillar has been connected by beams with the top of all its non adjacent pillars, then total number of beams = number of diagonals of 20 – sided polygon.
\(\because{ }^{20} C_2\) is selection of any two vertices of 20 – sided polygon which included the sides as well.
So, required number of total beams \(={ }^{20} C_2-20\)
[ \(\because\) the number of diagonals in a \(n\) – sided closed polygon \(={ }^n C_2-n\) ] \(=\frac{20 \times 19}{2}-20\)
\(=190-20=170\)

Hint

Given: ( 8 males, 5 females)
Committee to be selected = 11 members
\(\mathrm{m}=\) no. of ways the committee is formed with at least 6 males.
\(
\begin{aligned}
&\Rightarrow(6 M, 5 F) \text { or }(7 M, 4 F) \text { or }(8 M, 3 F) \\
&={ }^8 C_6 \times{ }^5 C_5+{ }^8 C_7 \times{ }^5 C_4+{ }^8 C_8 \times{ }^5 C_3=78
\end{aligned}
\)
\(n=\) no. of ways the committee is formed with atleast 3 female
\(
\begin{aligned}
&\Rightarrow(8 \mathrm{M}, 3 \mathrm{~F}) \text { or }(7 \mathrm{M}, 4 \mathrm{~F}) \text { or }(6 \mathrm{M}, 5 \mathrm{~F}) \\
&={ }^8 \mathrm{C}_8 \times{ }^5 \mathrm{C}_3+{ }^8 \mathrm{C}_7 \times{ }^5 \mathrm{C}_4+{ }^8 \mathrm{C}_6 \times{ }^5 \mathrm{C}_5 \\
&=10+40+28=78 \\
&\Rightarrow \mathrm{m}=\mathrm{n}=78
\end{aligned}
\)

Hint

Number of games played by men among themselves \(=2 \times \mathrm{m}_2=\mathrm{m}(\mathrm{m}-1)\)
Number of games played between men and women
\(
=2 \times{ }^m \mathrm{C}_1 \times{ }^2 \mathrm{C}_1=4 \mathrm{~m}
\)
\(
\begin{aligned}
&\therefore m(m-1)-4 m=84 \\
&\Rightarrow m^2-5 m-84=0 \\
&\Rightarrow m=12 \text { or }-7 \text { (Not possible) }
\end{aligned}
\)
Hence, \(m=12\)

Hint

Given :Girls- 5 and boys- 7
Team members:Girls- 2 and boys- 3
Total number of ways \(={ }^5 \mathrm{C}_2 \cdot{ }^7 \mathrm{C}_3\)
consider when \(\mathrm{A}\) and \(\mathrm{B}\) are always included \(={ }^5 \mathrm{C}_1{ }^5 \mathrm{C}_2\) as only 1 boy and 2 girls are to be selected
Require number of ways
\(=\) Total number of ways \(–\) When \(\mathrm{A}\) and \(\mathrm{B}\) are always included.
\(={ }^5 \mathrm{C}_2 \cdot{ }^7 \mathrm{C}_3-{ }^5 \mathrm{C}_1{ }^5 \mathrm{C}_2=300\).

Hint

Case (i): The first letter out of 5 can be chosen in 5 ways. The second letter can be chosen out of the remaining 4 in 4 ways. Again, the third letter can be chosen out of the remaining 3 in 3 ways, and lastly the fourth letter can be chosen out of the remaining 2 in 2 ways. Hence, the total arrangements of 4 lettered words with no repetition is \(5 \times 4 \times 3 \times 2=120\)

Case (ii): If two letters are \(\mathrm{R}\) and other two different letters are chosen from \(\mathrm{B}, \mathrm{A}, \mathrm{C}, \mathrm{K}\),
then the number of words \(={ }^4 \mathrm{C}_2 \times \frac{4 !}{2 !}=72\)
If two letters are A and other two different letters are chosen from B, R, C, K
then the number of words \(={ }^4 \mathrm{C}_2 \times \frac{4 !}{2 !}=72\)

Case (iii) : Similarly, when 2 letters are different and 2 letters are both \(\mathbf{R}\), the total arrangement is also 72 .
Case (iv) : Here, 2 letters are both \(\mathbf{R}\) and remaining 2 letters are both \(\mathbf{A}\) and \(h\) Hence, the total arrangement is \(4 ! / 2 ! \times 2 !=6\)
So, the total number of 4 lettered words is \(120+72+72+6=270\)

Hint

Out of 6 novels, 4 novels can be selected in \({ }^6 \mathrm{C}_4\) ways.
Also out of 3 dictionaries, 1 dictionary can be selected in \({ }^3 C_1\) ways.
Since the dictionary is fixed in the middle, we only have to arrange 4 novels which can be done in 4 ! ways.
Then the number of ways \(={ }^6 \mathrm{C}_4 \cdot{ }^3 \mathrm{C}_1 \cdot 4\) !
\(
=15 \times 3 \times 24=1080
\)

Hint

Given, \(\mathrm{X}\) has 7 friends, 4 of them are ladies and 3 are men while \(\mathrm{Y}\) has 7 friends, 3 of them ladies and 4 are men.

Possible cases for \(\mathrm{X}\) are
(1) 3 ladies, 0 man
(2) 2 ladies, 1 man
(3) 1 lady, 2 men
(4) 0 ladies, 3 men
Possible cases for \(\mathrm{Y}\) are
(1) 0 ladies, 3 men
(2) 1 lady, 2 men
(3) 2 ladies, 1 man
(4) 3 ladies, 0 man

Total number of required ways
\(
\begin{aligned}
&={ }^3 \mathrm{C}_3 \times{ }^4 \mathrm{C}_0 \times{ }^4 \mathrm{C}_0 \times{ }^3 \mathrm{C}_3+{ }^3 \mathrm{C}_2 \times{ }^4 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_1 \times{ }^4 \mathrm{C}_2 \times{ }^4 \mathrm{C}_2 \times{ }^3 \mathrm{C}_1+ \\
&{ }^3 \mathrm{C}_0 \times{ }^4 \mathrm{C}_3 \times{ }^4 \mathrm{C}_3 \times{ }^3 \mathrm{C}_0 \\
&=1+144+324+16=485
\end{aligned}
\)

Hint

Hint

The club consists of 6 girls and 4 boys. If a team of 4 members is to be selected which consists at most 1 boy (including 1 captain), then the number of ways of selecting the team is obtained as follows:
\(
{ }^4 \mathrm{C}_1\left({ }^4 \mathrm{C}_1 \cdot{ }^6 \mathrm{C}_3+{ }^6 \mathrm{C}_4\right)=4(80+15)=380 \text { ways }
\)

Hint

(c) Given
\(
n(\mathrm{~A})=4, n(\mathrm{~B})=2, n(\mathrm{~A} \times \mathrm{B})=8
\)
Required number of subsets
\(
\begin{aligned}
&={ }^8 \mathrm{C}_3+{ }^8 \mathrm{C}_4+\ldots .+{ }^8 \mathrm{C}_8=2^8-{ }^8 \mathrm{C}_0-{ }^8 \mathrm{C}_1-{ }^8 \mathrm{C}_2 \\
&=256-1-8-28=219
\end{aligned}
\)

Hint

(a) Number of diagonal \(=54\)
\(
\begin{aligned}
&\Rightarrow \frac{\mathrm{n}(\mathrm{n}-3)}{2}=54 \\
&\Rightarrow \mathrm{n}^2-3 \mathrm{n}-108=0 \Rightarrow \mathrm{n}^2-12 \mathrm{n}+9 \mathrm{n}-108=0 \\
&\Rightarrow \mathrm{n}(\mathrm{n}-12)+9(\mathrm{n}-12)=0 \\
&\quad \Rightarrow \mathrm{n}=12,-9 \Rightarrow \mathrm{n}=12(\mathrm{n} \neq-9)
\end{aligned}
\)

Hint

(c) Given
\(
n(\mathrm{~A})=2, n(\mathrm{~B})=4, n(\mathrm{~A} \times \mathrm{B})=8
\)
Required number of subsets \(=\)
\(
\begin{aligned}
&{ }^8 \mathrm{C}_3+{ }^8 \mathrm{C}_4+\ldots .+{ }^8 \mathrm{C}_8=2^8-{ }^8 \mathrm{C}_0-{ }^8 \mathrm{C}_1-{ }^8 \mathrm{C}_2 \\
&=256-1-8-28=219
\end{aligned}
\)

Hint

(b) We know,
\(
\begin{aligned}
&\mathrm{T}_n={ }^n \mathrm{C}_3, \mathrm{~T}_{n+1}={ }^{n+1} \mathrm{C}_3 \\
&\begin{array}{l}
\mathrm{ATQ}, \\
\mathrm{T}_{n+1}-\mathrm{T}_n={ }^{n+1} \mathrm{C}_3-{ }^n \mathrm{C}_3=10 \\
\Rightarrow \quad{ }^n \mathrm{C}_2=10 \quad \Rightarrow \quad n=5 .
\end{array}
\end{aligned}
\)

Hint

(b) Required number of triangles
\(
={ }^{12} \mathrm{C}_3-\left({ }^3 \mathrm{C}_3+{ }^4 \mathrm{C}_3+{ }^5 \mathrm{C}_3\right)=205
\)

Hint

(b) Each person gets at least one ball.
3 Persons can have 5 balls as follow.
\(
\begin{array}{|c|c|c|}
\hline \text { Person } & \text { No. of balls } & \text { No. of balls } \\
\hline \text { I } & 1 & 1 \\
\hline \text { II } & 1 & 2 \\
\hline \text { III } & 3 & 2 \\
\hline
\end{array}
\)
The number of ways to distribute balls \(1,1,3\) in first to three persons
\(
={ }^5 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_3
\)
Also 3 , persons having 1,1 and 3 balls can be arranged in \(\frac{3 !}{2 !}\) ways.
\(\therefore \quad\) Total no. of ways to distribute 1, 1, 3 balls to the three persons \(={ }^5 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_3 \times \frac{3 !}{2 !}=60\)
Similarly, total no. of ways to distribute 1, 2, 2 balls to three persons \(={ }^5 \mathrm{C}_1 \times{ }^4 \mathrm{C}_2 \times{ }^2 \mathrm{C}_2 \times \frac{3 !}{2 !}=90\)
\(\therefore \quad\) The required number of ways \(=60+90=150\)

Hint

If we see the blocks in terms of lines then there are \(2 \mathrm{~m}\) vertical lines and \(2 n\) horizontal lines. To form the required rectangle, we must select two horizontal lines, one even numbered (out of 2,4, \(.2 n\) ) and one odd numbered (out of \(1,3 \ldots . .2 n-1\) ) and similarly two vertical lines.
The number of rectangles
\(
={ }^m C_1 \cdot{ }^m C_1 \cdot{ }^n C_1 \cdot{ }^n C_1=m^2 n^2
\)

Hint

\(
\begin{aligned}
& T_n={ }^n C_3 ; T_{n+1}={ }^{n+1} C_3 \\
& \text { Now, } T_{n+1}-T_n=21 \Rightarrow{ }^{n+1} C_3-{ }^n C_3=21 \\
\Rightarrow & \frac{(n+1) n(n-1)}{3.2 .1}-\frac{n(n-1)(n-2)}{3.2 .1}=21 \\
\Rightarrow & n(n-1)(n+1-n+2)=126 \\
\Rightarrow & n(n-1)=42 \Rightarrow n(n-1)=7 \times 6, \quad \therefore n=7
\end{aligned}
\)

Hint

Two women can choose two chairs out of \(1,2,3,4\), in \({ }^4 C_2\) ways and can arrange themselves in 2! ways. Three men can choose 3
chairs out of 6 remaining chairs in \({ }^6 C_3\) ways and can arrange themselves in 3 ! ways
\(\therefore \quad\) Total number of possible arrangements are \({ }^4 C_2 \times 2 ! \times{ }^6 C_3 \times 3 !={ }^4 P_2 \times{ }^6 P_3\)

Alternate:

First the 2 women can select any chair of the 4 chairs numbered 1 to 4 in \({ }^4 P_2\) ways

Then there will be \(8-2=6\) chairs remaining, the 3 men can select any chair of the 6 chairs in \({ }^6 \mathrm{P}_3\) ways
So, total number of ways \(={ }^4 \mathrm{P}_2 \times{ }^6 \mathrm{P}_3\) ways

Hint

\(
\begin{aligned}
&\text { (c) }{ }^{47} C_4+\sum_{j=1}^5{ }^{52-j} C_3 \\
&={ }^{47} C_4+{ }^{51} C_3+{ }^{50} C_3+{ }^{49} C_3+{ }^{48} C_3+{ }^{47} C_3 \\
&={ }^{51} C_3+{ }^{50} C_3+{ }^{49} C_3+{ }^{48} C_3+\left({ }^{47} C_3+{ }^{47} C_4\right) \\
&={ }^{51} C_3+{ }^{50} C_3+{ }^{49} C_3+\left({ }^{48} C_3+{ }^{48} C_4\right) \\
&\quad\left[\mathrm{U} \operatorname{sing}{ }^n C_r+{ }^n C_{r+1}={ }^{n+1} C_{r+1}\right] \\
&={ }^{51} C_3+{ }^{50} C_3+\left({ }^{49} C_3+{ }^{49} C_4\right) \\
&={ }^{51} C_3+\left({ }^{50} C_3+{ }^{50} C_4\right)={ }^{51} C_3+{ }^{51} C_4={ }^{52} C_4
\end{aligned}
\)

Hint

(c) \({ }^n C_{r-1}=36,{ }^n C_r=84,{ }^n C_{r+1}=126\)
We know that
\(
\begin{aligned}
\frac{{ }^n C_{r-1}}{{ }^n C_r}=\frac{r}{n-r+1} \quad \Rightarrow \frac{36}{84}=\frac{r}{n-r+1} \\
\Rightarrow \frac{r}{n-r+1}=\frac{3}{7} \Rightarrow 3 n-10 r+3=0 \dots(i)\\
\text { Also, } \frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{r+1}{n-r}=\frac{84}{126}=\frac{2}{3} \\
\Rightarrow  2 n-5 r-3=0 \dots(ii)
\end{aligned}
\)
On solving (i) and (ii), we get \(n=9\) and \(r=3\).

Hint

Let \(n(A)=a, n(B)=b, n(A \cap B)=c\)
\(
\therefore \quad 1 \leq \mathrm{b}<\mathrm{a}
\)
Also given that \(\mathrm{A}\) and \(\mathrm{B}\) are independent events
\(
\therefore \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})
\)
\(
\begin{aligned}
&\Rightarrow \frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})} \times \frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})} \\
&\Rightarrow \frac{\mathrm{c}}{6}=\frac{\mathrm{a}}{6} \times \frac{\mathrm{b}}{6} \Rightarrow \mathrm{ab}=6 \mathrm{c}
\end{aligned}
\)
If \(a=6\) then \(b=c=5,4,3,2,1 \quad(b<a)\)
There is only one way to select all 6 elements of set A. Number of ways of selecting \(5,4,3,2\) or 1 elements in \(\mathrm{B}\) and \(\mathrm{A} \cap \mathrm{B}\) are
\(
{ }^6 \mathrm{C}_5+{ }^6 \mathrm{C}_4+{ }^6 \mathrm{C}_3+{ }^6 \mathrm{C}_2+{ }^6 \mathrm{C}_1=2^6-2=62
\)
If \(a=5\) then \(b=\frac{6 c}{5}\), which is not possible because ifc \(=5\) then \(b=6\), while \(\mathrm{b}<\mathrm{a}\).

If \(a=4\) then \(b=\frac{6 c}{4}=\frac{3 c}{2}\), which is possible because if \(c=2\) then \(b=3\)
2 elements in \(\mathrm{A} \cap \mathrm{B}\) can be selected in \({ }^6 \mathrm{C}_2\) ways.
2 additional elements in \(\mathrm{A}\) can be selected in \({ }^4 \mathrm{C}_2\) ways.
1 additional element in \(\mathrm{B}\) can be selected in \({ }^2 \mathrm{C}_1\) ways.
\(\therefore\) No. of ways for \(\mathrm{a}=4, \mathrm{~b}=3, \mathrm{c}=2\) are
\(
{ }^6 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1 \times{ }^2 \mathrm{C}_1=15 \times 6 \times 2=180
\)
If \(\mathrm{a}=3\) then \(\mathrm{b}=2 \mathrm{c} \Rightarrow \mathrm{c}=1, \mathrm{~b}=2\)
which can be done in \({ }^6 \mathrm{C}_1 \times{ }^5 \mathrm{C}_1+{ }^4 \mathrm{C}_2=6 \times 5 \times 6=180\) ways.
If \(a=2\) then \(b=3 c\) which is not possible
\(\therefore\) Total number of required ways \(=62+180+180=422\).

Hint

\(\mathrm{x}=10\) ! and \(\mathrm{y}={ }^{10} \mathrm{C}_1 \times{ }^9 \mathrm{C}_8 \times \frac{10 !}{2 !}=10 \times 9 \times \frac{10 !}{2 !}\)
\(
\therefore \quad \frac{y}{9 x}=\frac{10 \times 9 \times \frac{10 !}{2}}{9 \times 10 !}=5
\)

Hint

Here, \(-B_1 -B_2 -B_3 -B_4 -B_5\)
Out of 5 girls, 4 girls are together and 1 girl is separate. Now,
no of ways to select 2 positions out of 6 positions between boys \(={ }^6 \mathrm{C}_2 \dots(i)\)
4 girls are to be selected out of \(5={ }^5 \mathrm{C}_4\) ways …(ii)
Now, 2 groups of girls can be arranged in 2! ways. …(iii)
Also, the group of 4 girls and 5 boys is arranged in \(4 ! \times 5\) !ways
Now, total number of ways \(={ }^6 \mathrm{C}_2 \times{ }^5 \mathrm{C}_4 \times 2 ! \times 4 ! \times 5\) ! [from Eqs. (i), (ii), (iii) and (iv)]
\(\therefore \mathrm{m}={ }^6 \mathrm{C}_2 \times{ }^5 \mathrm{C}_4 \times 2 ! \times 4 ! \times 5\) !
And \(\mathrm{n}=5\) ! \(\times 6\) !
\(
\Rightarrow \frac{m}{n}=\frac{{ }^6 C_2 \times{ }^5 C_4 \times 2 ! \times 4 ! \times 5 !}{6 ! \times 5 !}=\frac{15 \times 5 \times 2 \times 4 !}{6 \times 5 \times 4 !}=5
\)

Hint

Number of adjacent lines \(=n\)
Number of non adjacent lines \(={ }^n C_2-n\)
\(
\begin{aligned}
&\therefore{ }^{\mathrm{n}} \mathrm{C}_2-n=n \Rightarrow \frac{n(n-1)}{2}-2 n=0 \\
&\Rightarrow n^2-5 n=0 \Rightarrow n=0 \text { or } 5 \text { But } n \geq 2 \Rightarrow n=5
\end{aligned}
\)

Hint

Given 8 vectors are
\((1,1,1),(-1,-1,-1) ;(-1,1,1),(1,-1,-1) ;(1,-1,1),(-1,1,-1) ;(1,1,-1)\), \((-1,-1,1)\)
These are 4 diagonals of a cube and their opposites.
For 3 non coplanar vectors first we select 3 groups of diagonals and its opposite in \({ }^4 C_3\) ways.
Then one vector from each group can be selected in \(2 \times 2 \times 2\) ways.
\(\therefore \quad\) Total ways \(={ }^4 \mathrm{C}_3 \times 2 \times 2 \times 2=32=2^5 \therefore p=5\)

Hint

For vowels not together
Number of ways to arrange \(\mathrm{L}, \mathrm{T}, \mathrm{T}, \mathrm{R}=\frac{4 !}{2 !}\)
Then put both \(\mathrm{E}\) in 5 gaps formed in \({ }^5 C_2\) ways.
\(
\therefore \text { No. of ways }=\frac{4 !}{2 !} \cdot{ }^5 C_2=120
\)

Hint

\(S \rightarrow 2, L \rightarrow 2, A, B, Y, U\).
\(\therefore\) Required number of ways \(={ }^2 C_1 \times{ }^5 C_2 \times \frac{4 !}{2 !}=240\).

Hint

Number of ways to select four questions from six questions \(={ }^6 C_4\)
And number of ways to answer these questions correctly \(=1 \times 1 \times 1 \times 1=1\)
And the number of ways to answer remain two questions wrongly \(=3 \times 3=9\)
\(
\begin{aligned}
&\therefore \text { Required number of ways }={ }^6 C_4 \times 1 \times 9 \\
&=\frac{6 !}{2 ! 4 !} \times 9 \\
&=\frac{6 \times 5}{2} \times 9=135
\end{aligned}
\)

Hint

Let \(x y z\) be the three digit number
\(
\begin{array}{ll}
x+y+z=10, x \leq 1, y \geq 0, z \geq 0 \\
x-1=t \Rightarrow x=1+t & x-1 \geq 0, t \geq 0 \\
t+y+z=10-1=9 & 0 \leq t, z, z \leq 9
\end{array}
\)
\(\therefore\) Total number of non-negative integral solution
\(
={ }^{9+3-1} C_{3-1}={ }^{11} C_2=\frac{11 \cdot 10}{2}=55
\)
But for \(t=9, x=10\), so required number of integers \(=55-1=54\).

Hint

\(
\begin{aligned}
\sum_{r=0}^{25}(4 r+1){ }^{25} C_r=4 \sum_{r=0}^{25} r \cdot{ }^{25} C_r+\sum_{r=0}^{25}{ }^{25} C_r \\
=4 \sum_{r=1}^{25} r \times \frac{25}{r}{ }^{24} C_{r-1}+2^{25}=100 \sum_{r=1}^{25}{ }^{24} C_{r-1}+2^{25} \\
=100.2^{24}+2^{25}=2^{25}(50+1)=51.2^{25}
\end{aligned}
\)
Hence, by comparison \(k=51\)

Hint

Total marbles are 5 red, 4 black and 3 white. 4 marbles are drawn, atmost 3 of them are red. Case are:
3R and 1 other \(\rightarrow^5 \mathrm{C}_3 \times{ }^7 \mathrm{C}_1=70\)
2Rand 2 other \(\rightarrow{ }^5 \mathrm{C}_2 \times{ }^7 \mathrm{C}_2=210\)
1R and 3 other \(\rightarrow{ }^5 \mathrm{C}_1 \times{ }^7 \mathrm{C}_3=175\)
Zero \(\mathrm{R}\) and 4 other \(\rightarrow^5 \mathrm{C}_0 \times{ }^7 \mathrm{C}_4=35\)
Total \(=490\)

Hint

Word “EXAMINATION” consists of 2A, 2I, 2N, E, X, M, T, O

Case I: All different letters are selected
Number of words formed: \({ }^8 \mathrm{C}_4 \times 4 !=1680\)

Case II: 2 letters are the same and 2 are different
Number of words formed \(={ }^3 \mathrm{C}_1 \times{ }^7 \mathrm{C}_2 \times \frac{4 !}{2 !}=756\)

Case III: 2 pairs of letters are the same
Number of words formed \(={ }^3 \mathrm{C}_2 \times \frac{4 !}{2 ! \times 2 !}=18\)
Total number of words formed \(=1680+756+18=2454\)

Hint

We know that total number of ways of selection of \(r\) days out of \(n\) days such that no two of them are consecutive \(=n-r+1_{C_r}\).
\(\therefore\) Selection of 4days out of 15 days such that no two of them are consecutive \(={ }^{15-4+1} C_4={ }^{12} C_4\)
\(
=\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2}=11 \times 5 \times 9=495
\)

Hint

Among \(6^{\prime}+’\) there are 7 intervals
\(
\mathrm{X}+\mathrm{X}+\mathrm{X}+\mathrm{X}+\mathrm{X}+\mathrm{X}+\mathrm{X}
\)
Number of ways of filling 7 gaps by \(4^{\prime}-^{\prime}={ }^7 \mathrm{C}_4=35\)
\(\therefore\) Required ways \(=35\)

Hint

We have in all 12 point. Since 3 points are used to form a triangle, therefore the total number of triangles including the triangles formed by collinear points on \(\mathrm{Ab}, \mathrm{BC}\), and \(\mathrm{CA}\) is \({ }^{12} \mathrm{C}_3=220\). But this includes the following: The number of triangles formed by 3 points on \(\mathrm{AB}={ }^3 \mathrm{C}_3=1\)
The number of triangles formed by 4 points on \(B C={ }^4 C_3=4\).
The number of triangles formed by 5 points on \(\mathrm{CA}={ }^5 \mathrm{C}_3=10\).
Hence, the required number of triangles
\(
=220-(10+4+1)=205 \text {. }
\)

Hint

\(
\begin{aligned}
&\text { (True) Consider, } \frac{(n+1)(n+2) \ldots(n+r)}{r !} \\
&=\frac{1.2 .3 \ldots(n-1) n(n+1)(n+2) \ldots(n+r)}{1.2 .3 \ldots n . r !} \\
&=\frac{(n+r) !}{n ! r !}={ }^{\mathrm{n}+\mathrm{r}} \mathrm{C}_{\mathrm{r}}=\text { An integral value } \\
&
\end{aligned}
\)
Thus given statement is true.

Hint

Hint

(b) \(\because a_n=\) number of all \(n\) digit \(+\) ve integers formed by the digits 0,1 or both such that no consecutive digits in them are 0 . and \(b_n=\) number of such \(n\) digit integers ending with 1
\(c_n=\) number of such \(n\) digit integers ending with 0
Clearly, \(a_n=b_n+c_n\left(\because a_n\right.\) can end with 0 or 1\()\)
Also \(b_n=a_{n-1}\)
and \(c_n=a_{n-2}[\because\) if last digit is 0 , second last has to be 1]
\(\therefore\) We get \(a_n=a_{n-1}+a_{n-2}, n \geq 3\)
Also \(a_1=1, a_2=2\),
Now by this recurring formula, we get
\(
\begin{aligned}
&a_3=a_2+a_1=3 \\
&a_4=a_3+a_2=3+2=5 \\
&a_5=a_4+a_3=5+3=8
\end{aligned}
\)
Also \(b_6=a_5=8\)

Hint

(a) By recurring formula, \(a_{17}=a_{16}+a_{15}\) is correct
Also \(c_{17} \neq c_{16}+c_{15}\)
\(
\begin{aligned}
&\Rightarrow a_{15} \neq a_{14}+a_{13} \quad\left(\because c_n=a_{n-2}\right) \\
&\therefore \text { Incorrect }
\end{aligned}
\)
Similarly, other parts are also incorrect.

Hint

Given that, there are 9 women and 8 men, a committee of 12 is to be formed including at least 5 women.
This can be doen in \(=(5\) women and 7 men \()+(6\) women and 6 men \()+(7\) women and 5 men \()+\) ( 8 women and 4 men \()+(9\) women and 3 men) ways
Total number of ways of forming committee
\(
\begin{aligned}
&=\left({ }^9 C_5 \cdot{ }^8 C_7\right)+\left({ }^9 C_6 \cdot{ }^8 C_6\right)+\left({ }^9 C_7 \cdot{ }^8 C_5\right)+\left({ }^9 C_8 \cdot{ }^8 C_4\right) \\
&+\left({ }^9 C_9 \cdot{ }^8 C_3\right) \\
&=1008+2352+2016+630+56=6062 \\
&
\end{aligned}
\)
(i) The women are in majority \(=2016+630+56=2702\)
(ii) The men are in majority \(=1008\) ways.

Hint

Out of 18 guests, 9 to be seated on side \(A\) and rest 9 on side \(B\). Now out of 18 guests, 4 particular guests desire to sit on one particular side say side \(A\) and other 3 on other side \(B\). Out of rest \(18-4-3=11\) guests we can select 5 more for side \(A\) and rest 6 can be seated on side \(B\). Selection of 5 out of 11 can be done in \({ }^{11} C_5\) ways and 9 guests on each side of the table can be seated in \(9 ! \times 9\) ! ways. Thus there are total \({ }^{11} C_5 \times 9\) ! \(\times 9\) ! arrangements.

Hint

Number of ways of drawing at least one black ball \(=1\) black and 2 other or 2 black and 1 other or 3 black \(={ }^3 C_1 \times{ }^6 C_2+{ }^3 C_2 \times{ }^6 C_1+{ }^3 C_3=3 \times 15+3 \times 6+1\) \(=45+18+1=64\)

Hint

The possible cases are:
Case I : A man invites (3 ladies) and woman invites (3 gentlemen)
\(
{ }^4 \mathrm{C}_3{ }^4 \mathrm{C}_3=16
\)

Case II : A man invites (2 ladies,1 gentleman) and woman invites (2 gentleman, 1 lady)
\(
\left({ }^4 \mathrm{C}_2{ }^3 \mathrm{C}_1\right)\left({ }^3 \mathrm{C}_1{ }^4 \mathrm{C}_2\right)=324
\)

Case III : A man invites (1 lady, 2 gentlemen) and woman invites (2 ladies, 1 gentleman)
\(
\Rightarrow\left({ }^4 \mathrm{C}_1{ }^3 \mathrm{C}_2\right)\left({ }^3 \mathrm{C}_2{ }^4 \mathrm{C}_1\right)=144
\)

Case IV : A man invites ( 3 gentlemen) and woman invites (3 ladies)
\(
{ }^3 \mathrm{C}_3{ }^3 \mathrm{C}_3=1
\)
Total number of ways
\(
=16+324+144+1=485
\)

Hint

The various possibilities to put 5 different balls in 3 different size boxes, when no box remains empty: The balls can be 1,1 and 3 in different or \(2,2,1\).
Case I: To put 1, 1 and 3 balls in different boxes. Selection of 1,1 and 3 balls out of 5 balls can be done in \({ }^5 C_1 \times{ }^4 C_1 \times{ }^3 C_3\) ways and then \(1,1,3\) can permute (as different size boxes) in 3 ! ways.
\(\therefore\) No. of ways
\(={ }^5 C_1 \times{ }^4 C_1 \times{ }^3 C_3 \times 3 !=5 \times 4 \times 1 \times 6=120\)

Case II: To put 2, 2 and 1 ball in different boxes. Selection of 2,2 and 1 balls out of 5 balls can be done in \({ }^5 C_2 \times{ }^3 C_2 \times{ }^1 C_1\) ways
And then 2, 2, 1 can permute (different boxes) in 3 ! Ways
\(\therefore\) No. of ways
\({ }^5 C_1 \times{ }^3 C_2 \times{ }^1 C_1 \times 3 !=10 \times 3 \times 1 \times 6=180\)
Combining case I and II, total number of required ways are \(=120+180=300\).

Hint

As all the X’s are identical, the question is of selection of 6 squares from 8 squares, so that no row remains empty. Here \(R_1\) has 2 squares, \(R_2\) has 4 squares, and \(\mathrm{R}_3\) has 2 squares. The selection scheme is as follows:
\(
\begin{array}{|c|c|c|}
\hline \mathrm{R}_1 & \mathrm{R}_2 & \mathrm{R}_3 \\
\hline 1 & 4 & 1 \\
\hline 1 & 3 & 2 \\
\hline 2 & 3 & 1 \\
\hline 2 & 2 & 2 \\
\hline
\end{array}
\)
Therefore, number of selection is
\(
\begin{aligned}
&{ }^2 \mathrm{C}_1 \times{ }^4 \mathrm{C}_4 \times{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_1 \times{ }^4 \mathrm{C}_3 \times{ }^2 \mathrm{C}_2+{ }^2 \mathrm{C}_2 \times{ }^4 \mathrm{C}_3 \times{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2 \times{ }^4 \mathrm{C}_2 \times{ }^2 \mathrm{C}_2=4+8+8+6=26 \\
\end{aligned}
\)

Hint

Number in this range will be 3 -digit number.
\(N=\overline{a b c}\) such that \(a+b+c=14\)
Also, \(a \geq 1, \quad a, b, c \in\{0,1,2, \ldots 9\}\)
Case I:
All 3-digit same
\(\Rightarrow 3 a=14\) not possible

Case II:
Exactly 2 digit same:
\(
\begin{aligned}
& \Rightarrow 2 a+c=14 \\
& (a, c) \in\{(3,8),(4,6),(5,4),(6,2),(7,0)\} \\
& \Rightarrow\left(\frac{3!}{2!}\right) \text { ways } \Rightarrow 5 \times 3-1 \\
& =15-1=14
\end{aligned}
\)

Case III:
All digits are distinct
\(
a+b+c=14
\)
without losing generality \(a>b>c\)
\(
\begin{aligned}
& (a, b, c) \in\left\{\begin{array}{l}
(9,5,0),(9,4,1),(9,3,2) \\
(8,6,0),(8,5,1),(8,4,2) \\
(7,6,1),(7,5,2),(7,4,3) \\
(6,5,3)
\end{array}\right. \\
& \Rightarrow 8 \times 3!+2(3!-2!)=48+8=56 \\
& =0+14+56=70 \\
&
\end{aligned}
\)

Hint

To solve this problem, we need to find the number of 3-digit numbers formed using the digits \(2,3,4,5\), and 7 , with no repetition of digits allowed, and these numbers should not be divisible by 3 . Let’s break down the solution step-by-step:
1. Calculating the total number of 3 -digit numbers without repetition:

The number of ways to form a 3-digit number from 5 unique digits \((2,3,4,5,7)\) without repetition can be calculated using permutations:
The total number of permutations for choosing 3 digits out of 5 and arranging them is given by:
\(
5 P 3=\frac{5!}{(5-3)!}=\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}=5 \times 4 \times 3=60
\)
So, there are 60 possible 3-digit numbers that can be formed from the digits 2, 3, 4, 5 , and 7 without repetition.

2. Finding the 3 -digit numbers divisible by 3 :

A number is divisible by 3 if the sum of its digits is divisible by 3 . Let’s consider the sums of every combination of these three digits to find out which sums are divisible by 3 :
Possible sums of combinations:
\(
\begin{aligned}
& 2+3+4=9 \text { (divisible by } 3 \text { ) } \\
& 2+4+7=13 \\
& 2+5+7=14 \\
& 2+3+5=10 \\
& 2+3+7=12 \text { (divisible by } 3 \text { ) } \\
& 3+4+5=12 \text { (divisible by } 3 \text { ) } \\
& 3+4+7=14 \\
& 3+5+7=15 \text { (divisible by } 3 \text { ) } \\
& 4+5+7=16
\end{aligned}
\)
The combinations whose sums are divisible by 3 are:
\(
\begin{aligned}
& 2,3,4 \\
& 2,3,7 \\
& 3,4,5 \\
& 3,5,7
\end{aligned}
\)
Since the sum of the digits is divisible by 3 for these combinations, any permutation of these sets will yield a number divisible by 3 :
The number of 3 -digit numbers that can be formed from each set of 3 digits is:
\(
3!=6
\)
So, the total number of 3-digit numbers divisible by 3 is:
4 sets \(\times 6\) permutations per set \(=24\)

3. Calculating the 3 -digit numbers not divisible by 3 :

To find the 3-digit numbers not divisible by 3 , we subtract the number of those divisible by 3 from the total number of 3 -digit numbers:
\(
60-24=36
\)
Therefore, the number of 3-digit numbers that can be formed using the digits 2,3, 4,5 , and 7 without repetition, and which are not divisible by 3 , is equal to 36 .

Hint

\(
\begin{aligned}
& \text { Number of ways }=\text { coefficient of } x^{16} \text { in }\left(x+x^2+\ldots+x^6\right)^4 \\
& =\text { coefficient of } x^{16} \text { in }\left(1-x^6\right)^4(1-x)^{-4} \\
& =\text { coefficient of } x^{16} \text { in }\left(1-4 x^6+6 x^{12} \ldots\right)(1-x)^{-4} \\
& ={ }^{15} C_3-4 \cdot{ }^9 C_3+6=125
\end{aligned}
\)

Hint

\(
\begin{array}{|l|l|l|}
\hline \text { Group A } & \text { Group B } & \text { Ways } \\
\hline 4 m & 4 w & { }^4 C_4 \cdot{ }^4 C_4=1 \\
3 m+1 w & 1 m+3 w & { }^4 C_1 \cdot{ }^5 C_1 \cdot{ }^5 C_1^4 C_3 \quad=400 \\
2 m+2 w & 2 m+2 w & { }^4 C_2 \cdot{ }^5 C_2{ }^5 C_2{ }^4 C_2 \quad=3600 \\
1 m+3 w & 3 m+w & { }^4 C_1{ }^5 C_3{ }^5 C_3{ }^4 C_1 \quad=1600 \\
4 w & 4 m & { }^5 C_4{ }^5 C_4 \quad=25 \\
\hline
\end{array}
\)

\(
\begin{aligned}
\text { Total ways } & =1+400+3600+1600+25 \\
& =5626
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
x+2 y+3 z=42 \\
x, y, z \geq 0 \\
\text { as } & \\
z=0 & x+2 y=42 \Rightarrow 22 \text { cases } \\
z=1 & x+2 y=39 \Rightarrow 20 \text { cases } \\
z=2 & x+2 y=36 \Rightarrow 19 \text { cases } \\
z=3 & x+2 y=33 \Rightarrow 17 \text { cases } \\
z=4 & x+2 y=30 \Rightarrow 16 \text { cases } \\
z=5 & x+2 y=27 \Rightarrow 14 \text { cases } \\
z=6 & x+2 y=24 \Rightarrow 13 \text { cases } \\
z=7 & x+2 y=21 \Rightarrow 11 \text { cases }
\end{array}
\)
\(
\begin{aligned}
& z=8 \quad x+2 y=18 \Rightarrow 10 \text { cases } \\
& z=9 \quad x+2 y=15 \Rightarrow 8 \text { cases } \\
& z=10 x+2 y=12 \Rightarrow 7 \text { cases } \\
& z=11 x+2 y=9 \Rightarrow 5 \text { cases } \\
& z=12 x+2 y=6 \Rightarrow 4 \text { cases } \\
& z=13 x+2 y=3 \Rightarrow 2 \text { cases } \\
& z=14 x+2 y=0 \Rightarrow 1 \text { case } \\
& \text { Total }=169 \\
&
\end{aligned}
\)

Hint

We have III, TT, D, S, R, B, U, O, N
Number of words with selection \(( a , a , a , b )\)
\(
={ }^8 C _1 \times \frac{4!}{3!}=32
\)
Number of words with selection \(( a , a , b , b )\)
\(
=\frac{4!}{2!2!}=6
\)
Number of words with selection \(( a , a , b , c )\)
\(
={ }^2 C _1 \times{ }^8 C _2 \times \frac{4!}{2!}=672
\)
Number of words with selection \(( a , b , c , d )\)
\(
\begin{aligned}
& ={ }^9 C _4 \times 4!=3024 \\
& \therefore \text { Total }=3024+672+6+32 \\
& =3734
\end{aligned}
\)

Hint

The problem involves choosing 15 questions out of a total of 20 available questions, with the constraint that at least 4 questions must be chosen from each of the three sections A, B, and C. To evaluate the total number of ways a student can select these questions, we need to consider every possible combination of questions from sections \(A, B\), and \(C\) that sum up to 15 questions while respecting the constraints.

Section \(A\) has 8 questions, Section \(B\) and \(C\) each have 6 questions. The student must choose at least 4 questions from each section, which satisfies the minimum requirement. However, since the student is to attempt a total of 15 questions, there are several combinations to consider, as outlined below:

Choosing 4 questions from \(A , 5\) from \(B\), and 6 from \(C\)
Choosing 4 questions from \(A, 6\) from \(B\), and 5 from \(C\)
Choosing 7 questions from \(A, 4\) from \(B\), and 4 from \(C\)
Choosing 6 questions from \(A , 5\) from \(B\), and 4 from \(C\)
Choosing 6 questions from \(A , 4\) from \(B\), and 5 from \(C\)
Choosing 5 questions from \(A , 5\) from \(B\), and 5 from \(C\)
Choosing 5 questions from \(A, 6\) from \(B\), and 4 from \(C\)
Choosing 5 questions from \(A , 4\) from \(B\), and 6 from \(C\)
\(
\begin{array}{|c|c|c|c|c|}
\hline A & B & C & \Rightarrow & \begin{array}{c}
\text { No. of } \\
\text { question }
\end{array} \\
\hline 4 & 5 & 6 & \rightarrow & { }^8 C_4{ }^6 C_5{ }^6 C_6 \\
\hline 4 & 6 & 5 & \rightarrow & { }^8 C_4{ }^6 C_6{ }^6 C_5 \\
\hline 7 & 4 & 4 & \rightarrow & { }^8 C_7{ }^6 C_4{ }^6 C_4 \\
\hline 6 & 5 & 4 & \rightarrow & { }^8 C_6{ }^6 C_5{ }^6 C_4 \\
\hline 6 & 4 & 5 & \rightarrow & { }^8 C_6{ }^6 C_4{ }^6 C_5 \\
\hline 5 & 5 & 5 & \rightarrow & { }^8 C_5{ }^6 C_5{ }^6 C_5 \\
\hline 5 & 6 & 4 & \rightarrow & { }^8 C_5{ }^6 C_6{ }^6 C_4 \\
\hline 5 & 4 & 6 & \rightarrow & { }^8 C_5{ }^6 C_4{ }^6 C_6 \\
\hline
\end{array}
\)
Total ways of select \(=11376\)

Hint

Words starting with \(E =360\)
Words starting with \(GE =60\)
Words starting with \(GN =60\)
Words starting with \(GTE =24\)
Words starting with \(GTN =24\)
Words starting with GTT \(=24\)
GTWENTY \(=1\)
Total \(=553\)

Hint

The word MATHEMATICS has
\(2 M\)
\(2 A\)
\(2 T\)
\(H, E, I, C, S\)

Case-I:

2 Alike 2 Alike 1 Diff
\({ }^3 C_2 \times{ }^6 C_1=18\)

Case-II:

2 Alike +3 Diff
\({ }^3 C_1 \times{ }^7 C_3=105\)

Case-III:

All different
\({ }^8 C_5=56\)
Total ways \(=179\)

Hint

\(
\begin{aligned}
& N =2310=231 \times 10 \\
& =3 \times 11 \times 7 \times 2 \times 5 \\
& A =\{2,3,5,7,11\} \\
& f(x)=\left[\log _2\left(x^2+\left[\frac{x^3}{5}\right]\right)\right] \\
& f (2)=\left[\log _2(5)\right]=2 \\
& f(3)=\left[\log _2(14)\right]=3 \\
& f(5)=\left[\log _2(25+25)\right]=5 \\
& f(7)=\left[\log _2(117)\right]=6 \\
& f(11)=\left[\log _2 387\right]=8
\end{aligned}
\)
\(
\text { Range of } f : B =\{2,3,5,6,8\}
\)
No. of one-one functions \(=5!=120\)

Hint

Word at \(315^{\text {th }}\) position
\(
\begin{aligned}
& \text { A….. }=5!=120 \\
& \text { G…… }=5!=120 \\
& \text { NA…. }=4!=24 \\
& \text { NG…. }=4!=24 \\
& \text { NP…. }=4!=24
\end{aligned}
\)
….. Till 312 words
\(
\begin{aligned}
& 313^{\text {th }} \text { word }=\text { NRAGPU } \\
& 314^{\text {th }} \text { word }=\text { NRAGUP } \\
& 315^{\text {th }} \text { word }=\text { NRAPGU }
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{n+1}{r+1} \times{ }^n C_r:{ }^n C_r: \frac{r}{n}{ }^n C_r=55: 35: 21 \\
& \Rightarrow \frac{n+1}{r+1}=\frac{55}{35} \text { and } \frac{n}{r}=\frac{35}{21}
\end{aligned}
\)
\(
\begin{gathered}
\Rightarrow 7 n-11 r=4 \dots(1)\\
3 n-5 r=0 \dots(2)
\end{gathered}
\)
Solving (1) and (2)
\(
\begin{aligned}
& r=6 \text { and } n=10 \\
& \Rightarrow 2 n+5 r=20+30=50
\end{aligned}
\)

Hint

Number of all possible triangles
\(=\) Number of selections of 3 points from 8 vertices
\(
={ }^8 C_3=56
\)
Number of triangle with one side common with octagon \(=8 \times 4=32\)
(Consider side \(A_1 A_2\). Since two points \(A_3, A_8\) are adjacent, 3 rd point should be chosen from remaining 4 points).
Number of triangles having two sides common with octagon: All such triangles have three consecutive vertices, viz., \(A_1 A_2 A_3, A_2 A_3 A_4 . \cdots A_8 A_1 A_2\).
Number of such triangles \(=8[latex]
[latex]\therefore\) Number of triangles with no side common
\(
=56-32-8=16 \text {. }
\)

Hint

Given set \(S=\left\{2^1, 2^2, \ldots 2^9\right\}\) which consist of 9 elements.
Maximum number of possible partitions (in set \(A, B\) and \(C\) )
\(
={ }^9 C_3 \cdot{ }^6 C_3 \cdot{ }^3 C_3=1680
\)

Alternate:

\(
\begin{aligned}
& \left\{2,2^2, 2^3 \ldots \ldots 2^9\right\} \\
& \Rightarrow \frac{9!}{3!3!3!3!} \times 3!=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{6 \times 6} \\
& =1680
\end{aligned}
\)

Hint

B,H,J,O
B. . . 4!=24
H. . . \(\frac{4!}{2!}=12\)
J. . . \(\frac{4!}{2!}=12\)

O B B H J \(\rightarrow 49{\text {th }}\)
O B B J H \(\rightarrow 50^{\text {th }}\) rank

Hint

Number of points on side \(A B=5\)
Number of points on side \(B C=6\)
Number of points on side \(A C=7\)
Number of ways selecting three points from side
\(
A B={ }^5 C_3
\)
Number of ways selecting three points from side
\(
B C={ }^6 C_3
\)
Number of ways selecting three points from side
\(
A C={ }^7 C_3
\)
Total number of triangle possible formed using the points \(P_1 P_2 \ldots P_{18}\)
\(
\begin{aligned}
& ={ }^{18} C_3-{ }^5 C_3-{ }^6 C_3-{ }^7 C_3 \\
& =816-10-20-35 \\
& =751
\end{aligned}
\)

Hint

Total ways to partition 5 into 4 parts are :
\((5,0,0,0) \Rightarrow 1\) way
\((4,1,0,0) \Rightarrow \frac{5!}{4!}=5\) ways
\((3,2,0,0) \Rightarrow \frac{5!}{3!2!}=10\) ways
\((2,2,0,1) \Rightarrow \frac{5!}{2!2!2!}=15\) ways
\((2,1,1,1) \Rightarrow \frac{5!}{2!(1!)^3 3!}=10\) ways
\((3,1,1,0) \Rightarrow \frac{5!}{3!2!}=10\) ways
Total \(\Rightarrow 1+5+10+15+10+10=51\) ways

Hint

After giving 2 apples to each child 15 apples left now 15 apples can be distributed in \({ }^{15+3-1} C _2={ }^{17} C _2\) ways
\(
\begin{aligned}
& =\frac{17 \times 16}{2} \\
& =136
\end{aligned}
\)

Explanation:

First, since each child must get at least 2 apples, let’s give 2 apples to each child right away. That accounts for 6 apples ( 2 apples for each of the 3 children). Now, we have \(21-6=15\) apples left to distribute freely among the three children.
The “stars and bars” technique involves representing the apples as stars (*) and the divisions between children as bars (|). For example, if we had 5 apples to distribute among three children, one possible distribution could be represented as \(* *|*| * *\). This means the first child gets 2 apples, the second child gets 1 apple, and the third child gets 2 apples.
In our case, we need to distribute 15 apples (stars) among the three children with 2 bars to create the partitions. We arrange 15 stars and 2 bars in a row, where the arrangement of stars and bars corresponds to a distribution of the apples.
The total number of objects we’re arranging is 15 apples +2 bars \(=17\) objects. We need to choose 2 positions out of these 17 to place the bars. The remaining positions will be occupied by the stars (apples).
The number of ways to choose 2 positions out of 17 for the bars is given by the binomial coefficient:
\(
\text { Number of ways }=\binom{17}{2}=\frac{17!}{2!(17-2)!}=\frac{17 \times 16}{2 \times 1}=136
\)

Hint

\(
\begin{aligned}
& { }^6 C _{ m }+2\left({ }^6 C _{ m +1}\right)+{ }^6 C _{ m +2}>{ }^8 C _3 \\
& { }^7 C _{ m +1}+{ }^7 C _{ m +2}>{ }^8 C _3 \\
& { }^8 C _{ m +2}>{ }^8 C _3 \\
& \therefore m =2 \\
& \text { And }{ }^{ n -1} P _3:{ }^n P _4=1: 8 \\
& \frac{( n -1)( n -2)( n -3)}{ n ( n -1)( n -2)( n -3)}=\frac{1}{8} \\
& \therefore n =8 \\
& \therefore{ }^{ n } P _{ m +1}+{ }^{ n +1} C _{ m }={ }^8 P _3+{ }^9 C _2 \\
& =8 \times 7 \times 6+\frac{9 \times 8}{2} \\
& =372
\end{aligned}
\)

Hint

3 Shelf empty: \((8,0,0,0) \rightarrow 1\) way
2 shelf empty: \(\left.\begin{array}{c}(7,1,0,0) \\ (6,2,0,0) \\ (5,3,0,0) \\ (4,4,0,0)\end{array}\right] \rightarrow 4\) ways
\(
\left.1 \text { shelf empty: } \begin{array}{rr}
(6,1,1,0) & (3,3,2,0) \\
(4,2,1,0) & (4,2,2,0)
\end{array}\right] \rightarrow 5 \text { ways }
\)
\(
\left.0 \text { Shelf empty : } \begin{array}{ll}
(1,2,3,2) & (5,1,1,1) \\
(2,2,2,2) & \\
(4,3,2,1,1) &
\end{array}\right] \rightarrow 5 \text { ways }
\)
\(
\text { Total }=15 \text { ways }
\)

Hint

\(
\begin{aligned}
& \alpha=\frac{(4!)!}{(4!)^{3!}}, \beta=\frac{(5!)!}{(5!)^{4!}} \\
& \alpha=\frac{(24)!}{(4!)^6}, \beta=\frac{(120)!}{(5!)^{24}}
\end{aligned}
\)
Let 24 distinct objects are divided into 6 groups of 4 objects in each group.
No. of ways of formation of group \(=\frac{24!}{(4!)^6 .6!} \in N\)
Similarly,
Let 120 distinct objects are divided into 24 groups of 5 objects in each group.
No. of ways of formation of groups
\(
=\frac{(120)!}{(5!)^{24} \cdot 24!} \in N
\)

Hint

Sum of first two digits
Sum of last two digits \(=\alpha\)
Case-I : \(\alpha=7\)
\(2 \times 12=24\) ways.

Hint

A number is divisible by 6 if it is divisible by both 2 and 3 . A number is divisible by 2 if its last digit is even, which means it must be either 2 or 4 from the given digits. A number is divisible by 3 if the sum of its digits is divisible by 3.
\(
\begin{aligned}
& (2,1,3),(2,3,4),(2,5,5),(2,2,5),(2,2,2) \\
& (4,1,1),(4,4,1),(4,4,4),(4,3,5) \\
& 2,1,3 \Rightarrow 312,132 \\
& 2,3,4 \Rightarrow 342,432,234,324 \\
& 2,5,5 \Rightarrow 552 \\
& 2,2,5 \Rightarrow 252,522 \\
& 2,2,2 \Rightarrow 222 \\
& 4,1,1 \Rightarrow 114 \\
& 4,4,1 \Rightarrow 414,144 \\
& 4,4,4 \Rightarrow 444 \\
& 4,3,5 \Rightarrow 354,534 \\
& \text { Total } 16 \text { numbers. }
\end{aligned}
\)

Hint

\(x_1+x_2+x_3+\ldots x_7=12\). This equation represents the number of ways to distribute 12 identical items (the sum of the digits) into 7 distinct boxes (the seven digits of the number), where each box can contain one of the numbers \(1,2,3\), or 4 .
Number of solutions
\(
\begin{aligned}
& =\text { Coefficient of } x^{12} \text { in }\left(x^1+x^2+x^3+x^4\right)^7 \\
& =\text { Coefficient of } x^5 \text { in }\left(1+x+x^2+x^3\right)^7 \\
& =\text { Coefficient of } x^5 \text { in }\left(1-x^4\right)^7(1-x)^{-7} \\
& =\text { Coefficient of } x^5 \text { in }\left(1-7 x^4\right)(1-x)^{-7} \\
& =\text { Coefficient of } x^5 \text { in }\left(1-7 x^4\right) \sum_{r=0}^{\infty}{ }^{7+r-1} C_r \cdot x^r \\
& ={ }^{11} C_5-7 \times{ }^7 C_1 \\
& =462-49=413
\end{aligned}
\)

Hint

The problem involves forming nine-digit numbers from three digits \(a, b, c\) which are in Arithmetic Progression (AP), used three times each, such that at least once, three consecutive digits are in AP.
We have the two possible sequences for the AP :
\(
\begin{aligned}
& a, b, c \\
& c, b, a
\end{aligned}
\)
This shows the flexibility in ordering the three digits that are in AP in our nine-digit number.
The next step is to choose the location of this sequence of three numbers within our nine-digit number.
Since there are nine places in our number and our sequence takes up three places, we have seven different starting points for our sequence : it can start at the first place, the second place, and so on, up to the seventh place.
Therefore, the number of ways to select 3 consecutive places out of the 9 places for the AP sequence is 7 .
However, we also have to account for the fact that our sequence can be in one of two orders ( \(a , b , c\) or \(c , b , a\) ). So, we multiply the number of starting points by 2 to get \({ }^7 C_1 \times 2=14\) ways to arrange the sequence within our nine-digit number.
The remaining 6 digits (two ‘a’, two ‘b’, two ‘c”) can be arranged in \(\frac{6!}{2!2!2!}\) ways.
Therefore, the total number of such nine-digit numbers is \({ }^7 C_1 \times 2 \times \frac{6!}{2!2!2!}=1260\)

Hint

This problem can be solved using the concept of derangements, which is a permutation of objects where no object appears in its original position. In this case, we have 5 students who should not sit in their allotted seats.
The formula for calculating the number of derangements (also known as subfactorials) is :
\(
D(n)=n!\left[\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\ldots \ldots .+(-1)^n \frac{1}{n!}\right]
\)
Where \(n\) is the number of students, in this case, 5 .
Using the formula, let’s calculate the derangements for 5 students :
\(
\begin{aligned}
& D(5)=5!\left(\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}\right) \\
& D(5)=120\left(1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}\right) \\
& D(5)=120\left(0+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}\right) \\
& D(5)=120\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}\right) \\
& D(5)=120\left(\frac{60}{120}-\frac{20}{120}+\frac{5}{120}-\frac{1}{120}\right) \\
& D(5)=120\left(\frac{44}{120}\right) \\
& D(5)=44
\end{aligned}
\)

Hint

We have, four digits are \(2,1,2,3\).
Total numbers when 1 is at unit digit \(=\frac{3!}{2!}=3\)
Total number when 2 is at unit digit \(=3!=6\)
Total numbers when 3 is at unit digit \(=\frac{3!}{2!}=3\)
Sum of digits at unit place \(=3 \times 1+6 \times 2+3 \times 3=24\)
\(\therefore\) Required sum \(=24 \times 1000+24 \times 100+24 \times 10+24 \times 1\)
\(=24 \times 1111=26664\)

Hint

Given that digits are \(1,2,3,4,5,6,7\)
Total permutations \(=7!\)
Let \(p=\) Number which containing string 153
\(q=\) Number which containing string 2467
\(
\begin{aligned}
& \therefore n(p)=5!\times 1 \\
& \Rightarrow n(q)=4!\times 1 \\
& \Rightarrow n(p \cap q)=2! \\
& \therefore n(p \cup q)=n(p)+n(q)-n(p \cap q) \\
& =5!+4!-2!=120+24-2=142
\end{aligned}
\)
\(\therefore n\) (neither string 143 nor string 2467)
\(
=7!-142=5040-142=4898
\)

Hint

Let, \(n\) be the total number of couples who participated in the tournament.
According to the question, \(2 \times{ }^n C_2 \times{ }^{n-2} C_2=840\)
\(
\begin{aligned}
& \Rightarrow{ }^n C_2 \times{ }^{n-2} C_2=420 \\
& \Rightarrow \frac{n!}{2!(n-2)!} \times \frac{(n-2)!}{(n-4)!2!}=420 \\
& \Rightarrow \frac{n(n-1)(n-2)(n-3)}{4}=420
\end{aligned}
\)
Put \(n=8\) satisfied the equation.
So, \(n=8\)
Hence, total number of players who participated in the tournament \(=2 \times 8=16\)

Hint

Given, word is UNIVERSE
Here, vowels are E, I, U and consonants are N, R, S, V
\(\therefore\) Required number of 4 -letters words, with or without meaning,
each consisting of 2 vowels and 2 consonants
\(
\begin{aligned}
& ={ }^3 C_2 \times{ }^4 C_2 \times 4! \\
& ={ }^3 C_1 \times{ }^4 C_2 \times 24 \\
& =3 \times 6 \times 24=432
\end{aligned}
\)

Hint

Total – (one child receive no orange + two child receive no orange)
\(
\begin{aligned}
& =3^{20}-\left({ }^3 C_1\left(2^{20}-2\right)+{ }^3 C_2 1^{20}\right) \\
& =3483638676
\end{aligned}
\)

Alternate:

Total ways without any restrictions :
There are \(3^{20}\) ways to distribute the oranges to the 3 children.
Number of ways one child receives no orange :
Choose 1 child out of the 3 to not receive any orange in \({ }^3 C_1=3\) ways. Distribute 20 oranges to the remaining 2 children in \(2^{20}\) ways. However, we’ve included the scenarios where the 2 children each get all the oranges. So, we subtract the 2 ways where one of the two remaining children gets all the oranges. \({ }^3 C_1\left(2^{20}-2\right)\)
Number of ways two children receive no orange :
Choose 2 children out of the 3 to not receive any oranges in \({ }^3 C_2=3\) ways. The third child will receive all 20 oranges in \(1^{20}=1\) way. \({ }^3 C_2 \times 1^{20}=3\)
Number of ways
\(
\begin{aligned}
& =\text { Total }-(\text { One child receive no orange }+ \text { two child receive no orange }) \\
& =3^{20}-\left({ }^3 C_1\left(2^{20}-2\right)+{ }^3 C_2 1^{20}\right) \\
& =3483638676
\end{aligned}
\)

Hint

\(
\begin{aligned}
& x+y+z=21 \\
& \because \quad x \geq 1, y \geq 3, y \geq 4 \\
& \therefore \quad x_1+y_1+z_1=13 \\
& \text { Number of solutions }={ }^{13+3-1} C_{3-1} \\
& ={ }^{15} C_2=\frac{15 \times 14}{2}=7 \times 15 \\
& =105
\end{aligned}
\)

Hint

A number will be divisible by 6 iff the digit at the unit place of the number is divisible by 2 and sum of all digits of the number is divisible by 3 .
Units, place must be occupied by 4 and hence, at least one 4 must be there.
Possible combination of \(4,5,9\) are as follows :
\(
\begin{aligned}
&\text { Possible combination of } 4,5,9 \text { are as follows : }\\
&\begin{array}{|l|l|l|l|}
\hline 4 & 5 & 9 & \begin{array}{l}
\text { Total number } \\
\text { of Number }
\end{array} \\
\hline 1 & 1 & 4 & \frac{5!}{4!}=5 \\
\hline 1 & 4 & 1 & \frac{5!}{4!}=5 \\
\hline 2 & 2 & 2 & \frac{5!}{2!2!}=30 \\
\hline 3 & 0 & 3 & \frac{5!}{2!3!}=10 \\
\hline 3 & 3 & 0 & \frac{5!}{2!3!}=10 \\
\hline 4 & 1 & 1 & \frac{5!}{3!}=20 \\
\hline 6 & 0 & 0 & \frac{5!}{5!}=1 \\
\hline
\end{array}
\end{aligned}
\)
\(
\text { Total }=5+5+30+10+10+20+1=81 \text {. }
\)

Hint

3 digit numbers divisible by either 2 or 3
\(P = n (\) divisible by 2\()+ n (\) divisible by 3\()- n (\) divisible by 6\()\)
\(P=450+300-150\)
\(P =600\)
\(Q = n (\) divisible by 14\()+ n (\) divisible by 21\()- n (\) divisible by 42\()\)
\(
=64+43-21=86
\)
3 digit number divisible by either 2 or 3
But not divisible by -1 so \(P – Q =600-86=514\)

Alternate:

\(
\begin{aligned}
& m (2)=450, n (3)=300, n (7)=128, n (2 \cap 7)=64 \\
& n (3 \cap 7)=43, n (2 \cap 3)=150, n (2 \cap 3 \cap 7)=21 \\
& \text { Total numbers }=450+300-150-(64+43-21) \\
& =514
\end{aligned}
\)

Explanation:

\(
\begin{aligned}
& A=\text { Numbers divisible by } 2 \\
& B=\text { Numbers divisible by } 3 \\
& C=\text { Numbers divisible by } 7 \\
& n(A \cup B)=n(A)+n(B)-n(A \cap B) \\
& =n(2)+n(3)-n(6) \\
& n(A)=n(2)=100,102 \ldots, 998=450 \\
& n(B)=n(3)=102,105, \ldots, 999=30 \\
& n(A \cap B)=n(6)=102,108, \ldots, 996=150 \\
& n(2 \text { or } 3)=450+300-150=600
\end{aligned}
\)
\(
\begin{aligned}
& n( A \cap C)=n(14)=112,126, \ldots, 994=64 \\
& n(A \cap B \cap C)=n(42)=126,168, \ldots, 966=21 \\
& n(B \cap C)=n(21)=105,126, \ldots \ldots, 987,=43 \\
& n(2 \text { or } 3 \text { not by } 7)=600-[64+43-21] =514
\end{aligned}
\)

Hint

Vowels : A,A,A,I,I,O
Consonants : S,S,S,S,N,N,T
\(\therefore\) Total number of ways in which vowels come together
\(
=\frac{8!}{4!2!} \times \frac{6!}{3!2!}=50400
\)

Hint

\(
A=\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{13} & a_{14}
\end{array}\right]
\)
Such that \(\Sigma a_{i i}=3,5,7\) or 11
Then for sum 3 , the possible entries are \((0,0,0,3),(0,0,1,2),(0,1,1,1)\).
Then total number of possible matrices \(=4+12+4=20\)
For sum 5 the possible entries are \((0,0,1,4),(0,0,2,3),(0,1,2,2),(0,1,1,3)\) and \((1,1,1,2)\).
\(\therefore\) Total possible matrices \(=12+12+12+12+4=52\)
For sum 7 the possible entries are \((0,0,3,4)\),
\((0,2,2,3),(0,1,2,4),(0,1,3,3),(1,2,2,2),(1,1,2,3)\) and \((1,1,1,4)\).
\(\therefore\) Total possible matrices \(=80\)
For sum 11 the possible entries are \((0,3,4,4)\), \((1,2,4,4),(2,3,3,3),(2,2,3,4)\).
\(\therefore\) Total number of matrices \(=52\)
\(\therefore\) Total required matrices \(=20+52+80+52=204\)