NCERT Exemplar MCQs

Hint

First the women choose the chairs from amongst the chairs numbered 1 to 4 .
Two women can be arranged in 4 chairs in \({ }^4 P_2\) ways.
In remaining 6 chairs, 3 men can be arranged in \({ }^6 P_3\) ways.
\(
\begin{aligned}
\therefore \text { Total number of possible arrangements } &={ }^4 P_2 \times{ }^6 P_3=\frac{4 !}{2 !} \times \frac{6 !}{3 !} \\
&=4 \times 3 \times 6 \times 5 \times 4=1440
\end{aligned}
\)

Hint

The alphabetical order of RACHIT is A, C, H, I, R and T
Number of words beginning with \(A=5\) !
Number of words beginning with \(C=5\) !
Number of words beginning with \(\mathrm{H}=5\) !
Number of words beginning with \(=5\) !
And Number of word beginning with R i.e. RACHIT \(=1\)
\(\therefore\) The rank of the word ‘RACHIT’ in the dictionary
\(
\begin{aligned}
&=5 !+5 !+5 !+5 !+1 \\
&=4 \times 5 !+1 \\
&=4 \times 5 \times 4 \times 3 \times 2 \times 1+1 \\
&=4 \times 120+1 \\
&=480+1 \\
&=481
\end{aligned}
\)

Hint

Number of question in part \(\mathrm{I}=6\)
Number of questions in part II \(=6\)
The different ways of doing the question are
i) Three from part I and 4 from part II
ii) 4 from part I and 3 from part II
iii) 5 from part I and 2 from part II
iv) 2 from part I and 5 from part II
\(
\begin{aligned}
&\Rightarrow\left({ }^6 \mathrm{C}_3 \times{ }^6 \mathrm{C}_4\right)+\left({ }^6 \mathrm{C}_4 \times{ }^6 \mathrm{C}_3\right)+\left({ }^6 \mathrm{C}_2 \times{ }^6 \mathrm{C}_5\right)+\left({ }^6 \mathrm{C}_5 \times{ }^6 \mathrm{C}_2\right) \\
&\Rightarrow(20 \times 15)+(15 \times 20)+(15 \times 6)+(6 \times 15) \\
&\Rightarrow 300+300+90+90 \\
&\Rightarrow 780
\end{aligned}
\)

Hint

Total number of points \(=18\)
Out of 18 numbers, 5 are collinear and we get a straight line by joining any two points.
\(\therefore\) Total number of straight-line formed by joining 2 points out of 18 points \(={ }^{18} C_2\).
Number of straight lines formed by joining 2 points out of 5 points \(={ }^5 \mathrm{C}_2\)
But 5 points are collinear and we get only one line when they are joined pairwise.
So, the required number of straight lines are
\(
\begin{aligned}
&={ }^{18} C_2-{ }^5 C_2+1 \\
&=\frac{18 \cdot 17}{2 \cdot 1}-\frac{5 \cdot 4}{2 \cdot 1}+1 \\
&=153-10+1 \\
&=144
\end{aligned}
\)
Hence, the total number of straight lines \(=144\)

Hint

Total number of persons \(=8\)
Number of person to be selected \(=6\)
It is given that, if \(\mathrm{A}\) is chosen then, \(\mathrm{B}\) must be chosen. Therefore, following cases arise:
Number of persons to be chosen out \(8=6\)
CASE I:When \(\mathrm{A}\) is chosen, \(B\) must be chosen
Number of ways of selecting 4 more persons from remaining 6 persons \(={ }^{8-2} C_{6-2}=6 C_4=15\)
CASE II:When \(\mathrm{A}\) is not chosen
Number of ways of selecting 6 persons from the remaining 7 persons= \(7 C_6=7\)
Total ways \(=15+7=22\)

Hint

Total number of persons \(=12\)
Number of persons to be selected \(=5\)
Out of 12 persons, a chairperson can be selected in \({ }^{12} C_1=12\) ways. The remaining 4 persons can be selected out of 11 persons in \({ }^{11} C_4=330\) ways.
\(\therefore\) Total number of committee \(=12 \times 330=3960\)

Hint

There are 26 English alphabets and 10 digits ( 0 to 9\()\).
It is given that each plate contains two different letters followed by three different digits.
Each plate contains 2 different letters followed by 3 different digits
Arrangement of 26 letters taken 2 at a time \(={ }^{26} P_2=26 \times 25=650\)
Arrangement of 10 digits taken 3 at a time \(={ }^{10} P_3=10 \times 9 \times 8=720\)
Total number of license plates \(=650 \times 720=468000\)

Hint

2 black balls can be selected from 5 black balls in \({ }^5 C_{2 \text { ways }}\) 3 red balls can be selected from 6 red balls in \({ }^6 C_3\) ways Total number of ways \(={ }^5 C_2 \times { }^6 C_3=10 \times 20=200\)

Hint

Total number of things \(=\mathrm{n}\)
3 things must be together (considered as 1 group)
\(\therefore\) The number of remaining things \(=n-3\)
Number of things to be selected \(=r\)
Out of \(\mathrm{r}, 3\) are always together
\(\therefore\) Number of ways of selection \(={ }^{n-3} C_{r-3}\)

Now arrangements of 3 things which are always taken together \(=3\) !
\(
\text { Number of arrangements of }(\mathrm{r}-3+1) \text { objects }=(\mathrm{r}-2) \text { ! }
\)

Number of arrangements of \((r-2)\) things \(=(r-2)\) !
\(\therefore\) Total number of arrangements \(={ }^{n-3} \mathrm{C}_{r-3} \times(r-2) ! \times 3\) !
Hence the required arrangements \(={ }^{n-3} C_{r-3} \times(r-2) ! \times 3\) !

Hint

Given word is: TRIANGLE Consonants are: T, R, N, G, L Vowels are: I, A, E Since we have to form words in such a way that no two vowels are together, we first arrange consonants. Five consonants can be arranged in 5 ! ways.
\(
\times C \times C \times C \times C \times C \times
\)
Arrangements of consonants (as shown above marked as \(C\) ) creates six gaps marked as ‘ \(\times\) ‘
Now three vowels can be arranged in any three of these 6 gaps in \({ }^6 P_3\) ways.
So, the total number of arrangements \(=5 ! \times{ }^6 P_3=120 \times 120=14400\)

Hint

We have to find the number of positive integers between 6000 and 7000 which are divisible by 5 . So, we have to fill up four vacant places _ _ _ _ by 10 different digits (from 0 to 9) satisfying the given conditions, and repetition of digits is not allowed.
The thousand’s place has to be filled up by the digit 6, so there is only one way to fill up this place.
As the numbers are divisible by 5 , so the unit’s place has to be filled up either by 5 or by 0 . Thus, there are only 2 ways to fill up the unit’s place.
Now, the hundred’s place can be filled up in 8 ways by any of the remaining 8 digits because at the thousand’s place and at the unit’s place two digits have already been used.
Similarly, the ten’s place can be filled up in 7 ways (out of 10 digits 3 has already been used)
\(\Rightarrow\) The number of required positive integers \(=1 \times 8 \times 7 \times 2=112\)

Hint

Given that \(P_1, P_2, P_3, \ldots P_{10}\) are 10 persons out of which 5 persons are to be arranged but \(P_1\) must occur and \(P_4\) and \(P_5\) never occur
\(\therefore\) Selection is to be done only for \(10-3=7\) persons
\(\therefore\) Number of selection \(={ }^7 C_4\)
\(=\frac{7 !}{4 !(7-4) !}\)
\(=\frac{7 !}{4 ! 3 !}\)
\(=\frac{7 \cdot 6 \cdot 5 \cdot 4 !}{4 ! \cdot 3 \cdot 2 \cdot 1}\)
\(=35\)
5 people can be arranged as 5 !
So, the number of arrangement \(=35 \times 5\) !
\(=35 \times 120\)
\(=4200\)
Hence, the required arrangement \(=4200\).

Hint

As 1 lamp is also enough for the hall to be illuminated.
Total number of ways \(={ }^{10} \mathrm{C}_1+{ }^{10} \mathrm{C}_2+\ldots .+{ }^{10} \mathrm{C}_{10}\)
\(
\begin{aligned}
&\Rightarrow 2^{10}-1 \\
&\Rightarrow 1024-1 \\
&\Rightarrow 1023
\end{aligned}
\)

Hint

Number of ways of selecting 3 balls out of 9 balls without any restriction is \({ }^9 C_3\)
Number of ways of selecting 3 balls such that no black ball is selected is \({ }^6 C_3\)
Number of ways such that at least one black ball is selected is \({ }^9 C_3-{ }^6 C_3=84-20=64\)

Hint

\(
\begin{aligned}
&\frac{n_{C_r}}{n_{C_{r-1}}}=\frac{n-r+1}{r}=\frac{84}{36}=\frac{7}{3} \\
&3 n-3 r+3=7 r \\
&10 r-3 n=3 \ldots \ldots \ldots(i) \\
&\frac{n_{C_{r+1}}}{n_{C_r}}=\frac{n-(r+1)+1}{r+1}=\frac{126}{84} \\
&\frac{n-r}{r+1}=\frac{3}{2} \\
&2 n-2 r=3 \lambda-q^5 . \\
&2 n-5 r=3 \ldots \ldots \text { (ii) } \\
&\text { Equation (i) }+2 \text { Equation(ii) } \\
&10 r-3 \mathrm{n}+4 \mathrm{n}-10 \mathrm{r}=3+6 \\
&\mathrm{n}=9 \\
&\mathrm{r}=3 \\
&{ }^r C_2={ }^3 C_2=3
\end{aligned}
\)

Hint

As per the question,
The digits that can be used \(=3,5,7,8,9\)
As, no digits can be repeated,
No. of integers is \({ }^5 \mathrm{P}_5=5\) ! \(=120\)
For a 4 digit integer to be greater than 7000 ,
The 4-digit integer should begin with 7,8 or 9.
The no. of such integer \(=3 \times{ }^4 \mathrm{P}_3=3 \times{ }^4 \mathrm{P}_3=3(24)=72\)
As a result, the total number of ways \(=120+72=192\)

Hint

Sol: It is given that no two lines are parallel which means that all the lines are intersecting and no three lines are concurrent.
One point of intersection is created by two straight lines.
Number of points of intersection = Number of combinations of 20 straight lines taken two at a time
\(
={ }^{20} C_2=\frac{20 !}{2 ! 18 !}=\frac{20 \times 19 \text { Solutions }}{2 \times 1}=190
\)

Hint

Suppose that the first two digits are 41, the remaining 4 are to be chosen from \(0,2,3,5,6,7,8,9\) so as to make all the digits distinct.
Thus remaining 4 can be chosen in \({ }^8 \mathrm{P}_4\) ways = \(
{ }^8 p_4=8 \times 7 \times 6 \times 5=1680 \text { ways. }
\).
Hence, the number of 2 digits that the telephone number begins with =5
First, two digits can be filled in 5 ways.
Therefore, the number of telephone numbers having six distinct digits \(=5 \times 1680\)
\(=8400\)

Hint

It is given that 2 questions are compulsory out of 5 questions. So, the other 2 questions can be selected from the remaining 3 questions in \({ }^3 C_2=3\) ways.

Hint

\(
\begin{aligned}
&\text { No. of diagonals in a polygon of } \mathrm{n} \text { sides }=\frac{\mathrm{n}(\mathrm{n}-3)}{2} \\
&\frac{\mathrm{n}(\mathrm{n}-3)}{2}=44 \\
&\mathrm{n}(\mathrm{n}-3)=88 \\
&\mathrm{n}(\mathrm{n}-3)=11(11-3) \\
&\therefore \mathrm{n}=11 \\
&\therefore \text { No. of sides }=11 .
\end{aligned}
\)

Alternate:

It is known that,
\(
\begin{aligned}
&{ }^n C_r \\
&=\frac{n !}{r !(n-r) !}
\end{aligned}
\)
Let’s suppose the no. of sides the given polygon have \(=\mathrm{n}\)
So now,
The no. of line segments obtained by joining \(n\) vertices \(={ }^n C_2\)
Therefore, the no. of diagonals of the polygon \(={ }^{\mathrm{n}} \mathrm{C}_2-\mathrm{n}=44\)
\(
\begin{aligned}
&\frac{n(n-1)}{2}-n=44 \\
&n^2-3 n-88=0 \\
&(n-11)(n+8)=0 \\
&n=11 \text { or } n=-8
\end{aligned}
\)
As a result, the polygon has 11 sides.

Hint

Number of mice \(=18\),
Number of groups \(=3\)
Since the groups are equally large,
The number of mice in each group can be \(=6\) mice
The number of ways of placement of mice \(=18\) !
For each group the placement of mice \(=6\) !
Hence, the required number of ways \(=18 ! /(6 ! 6 ! 6 !)\)
\(
=18 ! /(6 !)^3
\)

Hint

Total number of marbles = 6 white + 5 red = 11 marbles

(a) If they can be of any colour means we have to select 4 marbles out of 11
\(\therefore\) Required number of ways \(={ }^{11} \mathrm{C}_4\)

(b) Two white marbles can be selected in \({ }^6 \mathrm{C}_2\)
Two red marbles can be selected in \({ }^5 \mathrm{C}_2\) ways.
\(\therefore\) Total number of ways \(={ }^6 \mathrm{C}_2 \times{ }^5 \mathrm{C}_2=15 \times 10=150\)

(c) If they all must be of same colour,
Four white marbles out of 6 can be selected in \({ }^6 \mathrm{C}_4\) ways.
And 4 red marbles out of 5 can be selected in \({ }^5 \mathrm{C}_4\) ways.
\(\therefore\) Required number of ways \(={ }^6 \mathrm{C}_4+{ }^5 \mathrm{C}_4=15+5=20\)

Hint

Total number of players \(=16\)
We have to select a team of 11 players
So, number of ways \(={ }^{16} \mathrm{C}_{11}\)
(i) If two particular players are included then more 9 players can be selected from the remaining 14 players in \({ }^{14} \mathrm{C}_9\)
(ii) If two particular players are excluded then all 11 players can be selected from the remaining 14 players in \({ }^{14} C_{11}\)

Hint

Number of students in sports team \(=11\)
Class XI and XII have 20 students
At least 5 students should be selected from each class for the sports team.
\(\therefore\) Number of ways the team can be formed \(=6\) from XI and 5 from XII or 6 from XII and 5 from XI
\(
\text { Required number of ways }={ }^{20} C_5 \times{ }^{20} C_6+{ }^{20} C_6 \times{ }^{20} C_5=2 \times{ }^{20} C_5 \times{ }^{20} C_6
\)

Hint

No of girls \(=4\)
No of boys \(=7\)
Total member to be selected \(=5\)
(i)
No girl
No of girls to be selected \(=0\)
No of boys to be selected \(=5\)
No of ways in which girl can be selected \(={ }^4 \mathrm{C}_0\)
No of ways in which boys can be selected \(={ }^7 \mathrm{C}_5\)
No of ways team member can be selected with no girls \(={ }^4 \mathrm{C}_0 \times{ }^7 \mathrm{C}_5\)
\(
=\frac{7 !}{5 ! 2 !}=21
\)
(ii)
Atleast one boys and one girl
No of girls to be selected \(=1,2,3,4\)
No of boys to be selected \(=1,2,3,4\)
No of ways if there are 1 girl and 4 boys \(={ }^4 \mathrm{C}_1{ }^7 \mathrm{C}_4\)
No of ways if there are 2 girls and 3 boys \(={ }^4 \mathrm{C}_2{ }^7 \mathrm{C}_3\)
No of ways if there are 3 girls and 2 boys \(={ }^4 \mathrm{C}_3{ }^7 \mathrm{C}_2\)
No of ways of there are 4 girls and 1 boys \(={ }^4 \mathrm{C}_4{ }^7 \mathrm{C}_1\)
Total no of ways \(={ }^4 \mathrm{C}_1{ }^7 \mathrm{C}_4+{ }^4 \mathrm{C}_2{ }^7 \mathrm{C}_3+{ }^4 \mathrm{C}_3{ }^7 \mathrm{C}_2+{ }^4 \mathrm{C}_4{ }^7 \mathrm{C}_1\)
\(=4 \times 35+6 \times 35+4 \times 21+7\)
\(=140+210+84+7\)
\(=441\)
(iii)
Atleast 3 girls
No of girls to be selected \(=3\) or 4
No of boys to be selected \(=2\) or 1
Total is ways \(={ }^4 \mathrm{C}_3{ }^7 \mathrm{C}_2+{ }^4 \mathrm{C}_4{ }^7 \mathrm{C}_1\)
\(
=4 \times 21+7=84+7=91
\)

Hint

(a) We have, \({ }^n C_{12}={ }^n C_8\)
\(
\begin{aligned}
&\Rightarrow \quad{ }^n C_{n-12}={ }^n C_8 \quad\left[\because{ }^n C_r={ }^n C_{n-r}\right] \\
&\Rightarrow \quad n-12=8 \Rightarrow n=20 \\
&
\end{aligned}
\)

Hint

(b) Number of outcomes when a coin tossed \(=2(\) Head or Tail \()\)
\(\therefore\) Total possible outcomes when a coin tossed 6 times \(=2 \times 2 \times 2 \times 2 \times 2 \times 2=64\)

Hint

(c) Given digits 2,3,4 and 7, we have to form four-digit numbers using these digits. \(\therefore\) Required number of ways \(={ }^4 P_4=4 !=4 \times 3 \times 2 \times 1=24\)

Hint

(b) If the unit place is ‘ 3 ‘ then remaining three places can be filled in 3 ! ways. Thus ‘ 3 ‘ appears in unit place in 3 ! times.
Similarly each digit appear in unit place \(3 !\) times.
So, sum of digits in unit place \(=3 !(3+4+5+6)=18 \times 6=108\)

Explanation:

Let 3 be the unit place number, then there are 3 choices for 1 oth place number, as the 10th place digit can be chose from 4,5 , 6 . Similarly there are 2 choices form 100th place digit and 1 choice for 1000 th place digit.
Hence there are \(3 \cdot 2 \cdot 1=6\) numbers ending with 3 . These 6 number will add \(6 \times 3=18\) to the sum of the digits in the unit’s place.

Similarly, there will be 6 numbers each ending with 4,5 and 6 and they will add \(4 \cdot 6=24 ; 5 \cdot 6=30\) and \(6 \cdot 6=36\) to the sum.

Hence, the sum of the digits in the unit’s place of all numbers formed with the help of \(3,4,5,6\) taken all at a times is \(18+24+30+36=108\).

Hint

(c) Given number of vowels \(=4\) and number of consonants \(=5\) We have to form words by 2 vowels and 3 consonants.
So, let’s first select 2 vowels and 3 consonants.
Number of ways of selection \(={ }^4 \mathrm{C}_2 \times{ }^5 \mathrm{C}_3=6 \times 10=60\) Now, these letters can be arranged in 5 ! ways.
So, total number of words \(=60 \times 5 !=60 \times 120=7200\)

Hint

The sum of the numerals \(0,1,2,3,4\) and 5 is 15 .
We know that a five digit number is divisible by 3 if and only if the sum of its digits by 3 .
Therefore, we should not use either 0 or 3 while forming the five digit numbers.
If we do not use 0 , then the remaining digits can arranged in \({ }^5 \mathrm{P}_5=5 !=120\) ways.

If we do not use 3, then the remaining digits can be arranged in \({ }^5 \mathrm{P}_5-{ }^4 \mathrm{P}_4=5\) ! – \(4 !=120-24=96\) ways to obtain a five-digit number.
The total number of 5-digit number is \(120+96=216\).

Note: If digits 0, 1,2,4, and 5 are used then first place from left can be filled in 4 ways and the remaining 4 places can be filled in 4! ways. So in this case required numbers are 4 x 4! ways.

Hint

Step 1. Find the total number of persons in the room:
Total handshakes \(=66\)
Let there are \(n\) people in the room. There should be two hands involved for a handshake.
\(\therefore\) total number of persons in a room are:
\(
\begin{aligned}
&{ }^n C_2=66 \\
&\Rightarrow \frac{{ }^{n(n-1)}}{2}=66 \\
&\Rightarrow n^2-n-132=0
\end{aligned}
\)
Step 2. Solve the equation for \(\mathrm{n}\), we get
\(
\begin{aligned}
&\Rightarrow n^2-12 n+11 n-132=0 \\
&\Rightarrow n(n-12)+11(n-12)=0 \\
&\Rightarrow(n-12)(n+11)=0 \\
&\therefore n=12 \text { and } n=-11
\end{aligned}
\)
The value of \(n\) cannot be negative.
\(
\therefore n=12
\)
Thus, The number of persons in the room are \(\mathbf{1 2}\)

Hint

Total number of triangles formed from 12 points taking 3 at a time \({ }^{12} \mathrm{C}_3\)
But given that out of 12 points, 7 are collinear
So, these seven points will form no triangle (any three points selected from given seven collinear points does not form triangle.)
\(\therefore\) The required number of triangles \(={ }^{12} C_3-{ }^7 C_3\)
\(=\frac{12 !}{3 ! 9 !}-\frac{7 !}{3 ! 4 !}\)
\(=\frac{12 \times 11 \times 10 \times 9 !}{3 \times 2 \times 1 \times 9 !}-\frac{7 \times 6 \times 5 \times 4 !}{3 \times 2 \times 1 \times 4 !}\)
\(=\frac{12 \times 11 \times 10}{3 \times 2}-\frac{7 \times 6 \times 5}{3 \times 2}\)
\(=220-35\)
\(=185\)

Hint

(b) To form parallelogram we required a pair of line from a set of 4 lines and another pair of line from another set of 3 lines.
Required number of parallelograms \(={ }^4 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2=6 \times 3=18\)

Hint

Total number of players \(=22\)
We need to select 11 players
Given condition says,
Exclude 4 particular player gives us 18 players available and including 2 particular players.
Again it said that 2 out of 18 has to be included in the squad, so we are left with 9 selection out of the remaining 16 players.
This implies only 9 need to be selected out of the remaining 16 (22-2-4=16).
Hence, the number of ways of selection is \({ }^{16} \mathrm{C}_9\).

Hint

Number of 5 -digit numbers when all the digits are repeated \(=10 \times 10 \times 10 \times 10 \times 10=100000\)
Numbers without the repeated digits \(=10 \times 9 \times 8 \times 7 \times 6=30240\)
Therefore, numbers having at least one of their digits repeated is \(=100000-30240=69760\)
Hence, the correct option is (d).

Hint

(a) Number of men = 4; Number of women \(=6\)
It is given that the committee includes at least two men and exactly twice as many women as men.
So, we can select either 2 men and 4 women or 3 men and 6 women.
\(\therefore\) Required number of committee formed \(={ }^4 \mathrm{C}_2 \times{ }^6 \mathrm{C}_4+{ }^4 \mathrm{C}_3 \times{ }^6 \mathrm{C}_6\)
\(=6 \times 15+4 \times 1=94\)

Hint

(c) We have to form 9-digit number which has all different digit.

\(
\text { There are } 10 \text { different digits. Those are } 0,1,2,3,4,5,6,7,8,9 \text {. }
\)
First digit from the left can be filled in 9 ways (excluding ‘ 0 ‘).
Now nine digits are left including ‘ \(O\) ‘.
So remaining eight places can be filled with these nine digits in \({ }^9 P_8\) ways.
So, total number of numbers \(=9 \times{ }^9 P_8=9 \times 9\) !

Hint

Total number of letters in the ‘ARTICLE’ is 7 out which \(\mathrm{A}, \mathrm{E}, \mathrm{I}\) are vowels and \(\mathrm{R}, \mathrm{T}, \mathrm{C}, \mathrm{L}\) are consonants Given that vowels occupy even place
\(\therefore\) Possible arrangement can be shown as below
C, V, C, V, C, V, C i.e. on \(2^{\text {nd }}, 4^{\text {th }}\) and \(6^{\text {th }}\) places
Therefore, number of arrangement \(={ }^3 P_3=3 !=6\) ways
Now consonants can be placed at \(1,3,5\) and \(7^{\text {th }}\) place
\(\therefore\) Number of arrangement \(={ }^4 P_4=4 !=24\)
So, the total number of arrangements \(=6 \times 24=144\).

Hint

Possible number of choosing 5 different green dyes \(=2^5\)
Possible number of choosing 4 blue dyes \(=2^4\)
And possible number of choosing 3 red dyes \(=2^3\)
If atleast one blue and one green dyes are selected then the total number of selection
\(
\begin{aligned}
&=\left(2^5-1\right) \times\left(2^4-1\right) \times 2^3 \\
&=31 \times 15 \times 8 \\
&=3720
\end{aligned}
\)

Hint

We have, \({ }^n P_r=840\) and \({ }^n C_r=35\)
Now \({ }^n P_r={ }^n C_r \cdot r !\)
\(
\Rightarrow \quad 840=35 \times r ! \Rightarrow r !=24 \quad \therefore r=4
\)

Hint

\(
\begin{aligned}
{ }^{15} C_8+{ }^{15} C_9-{ }^{15} C_6-{ }^{15} C_7 &={ }^{15} C_8+{ }^{15} C_9-{ }^{15} C_{15-6}-{ }^{15} C_{15-7}\left[\because{ }^n C_r={ }^n C_{n-r}\right] \\
&={ }^{15} C_8+{ }^{15} C_9-{ }^{15} C_9-{ }^{15} C_8=0
\end{aligned}
\)

Hint

Letters of the word INTERMEDIATE are: Vowels (I, E, E, I, A, E) and consonants (N, T, R, M, D, T)
Now we have to arrange these letters if no two vowels come together.
So, first, arrange six consonants in \(\frac{6 !}{2 !}\) ways.
The arrangement of six consonants creates seven gaps.
Six vowels can be arranged in these gaps in \({ }^7 C_6 \times \frac{6 !}{2 ! 3 !}\) ways.
So, total number of words \(=\frac{6 !}{2 !} \times{ }^7 C_6 \times \frac{6 !}{2 ! 3 !}=360 \times 7 \times 60=151200\)

[Hint: Number of ways of arranging 6 consonants of which two are alike is \(\frac{6 !}{2 !}\) and number of ways of arranging vowels \(={ }^7 \mathrm{P}_6 \times \frac{1}{3 !} \times \frac{1}{2 !}\). \(]\)

Hint

The possible selection can be either ‘two red and one other than red’ or ‘three red’.
So, the required number of ways \(={ }^5 C_2 \times{ }^7 C_1+{ }^5 C_3=10 \times 7+10=80\)

Hint

For a six-digit number; possible options of digit are \(1,3,5,7,9\) Since each place has all 5 options available.

\(\therefore\) number of six-digit numbers, all digits of which are odd is
\(
\begin{aligned}
&5 \times 5 \times 5 \times 5 \times 5 \times 5 \\
&=56
\end{aligned}
\)

Hint

Let the number of participating teams be \(\mathrm{n}\)
Given that every two teams played one match with each other.
\(\therefore\) Total number of matches played \({ }^n C_2\)
So \({ }^n C_2=153\)
\(
\begin{aligned}
&\Rightarrow \frac{n(n-1)}{2}=153 \\
&\Rightarrow n^2-n=306 \\
&\Rightarrow n^2-n-306=0 \\
&\Rightarrow n^2-18 n+17 n-306=0 \\
&\Rightarrow n(n-18)+17(n-18)=0 \\
&\Rightarrow(n-18)(n+17)=0 \\
&\Rightarrow n-18=0 \text { and } n+17=0 \\
&\Rightarrow n=18, n \neq-17
\end{aligned}
\)
Hence, the value of the filler is 18 .

Hint

First arrange six ‘+’ signs.
Number of ways of arrangement is ‘ 1 ‘ (as signs are identical).
Now, seven gaps are created from which four are to be chosen to put four ‘-‘ signs.
Four gaps can be selected in \({ }^7 C_4\) ways.
In these four gaps, ‘-‘ can be arranged in only one way as signs are identical.
\(\therefore\) Total number of ways \(=1 \times{ }^7 C_4 \times 1=35\)

Hint

Number of men \(=10 ; \quad\) Number of women \(=7\)
We have to form a committee of 6 persons containing at least 3 men and 2 women containing at least 3 men and 2 women.
\(\therefore\) Number of ways \(={ }^{10} C_3 \times{ }^7 C_3+{ }^{10} C_4 \times{ }^7 C_2=120 \times 35+210 \times 21=8610\)
Total number of committee when two particular women are never together
\(=\) Total number of ways \(–\) Number of ways when two particular women are together
\(
\begin{aligned}
&=\left({ }^{10} C_3 \times{ }^7 C_3+{ }^{10} C_4 \times{ }^7 C_2\right)-\left({ }^{10} C_4+{ }^{10} C_3 \times{ }^5 C_1\right) \\
&=(120 \times 35+210 \times 21)-(210+120 \times 5) \\
&=4200+4410-(210+600)=7800
\end{aligned}
\)

[Hint: At least 3 men and 2 women: The number of ways \(={ }^{10} \mathrm{C}_3 \times{ }^7 \mathrm{C}_3+{ }^{10} \mathrm{C}_4 \times{ }^7 \mathrm{C}_2\). For 2 particular women to be always there: the number of ways \(={ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_3 \times{ }^5 \mathrm{C}_1\). The total number of committees when two particular women are never together \(=\) Total \(–\) together. \(]\)

Hint

Required number of ways such that 3 balls are drawn and at least one black ball is included in the draw \(={ }^3 C_1 \times{ }^6 C_2+{ }^3 C_2 \times{ }^6 C_1+{ }^3 C_3\)
\(
=3 \times 15+3 \times 6+1=45+18+1=64
\)

Alternate Solution:

Step 1: Find the number of ways of selecting 3 balls
We have been given that, a box contains 2 white balls, 3 black balls and 4 red balls.
We need to find the number of ways 3 balls can be drawn from the box, if atleast one black ball is to be included in the draw.
Since there are 2 white balls, 3 black balls and 4 red balls in a box,
Total balls \(=9\)
The number of ways of selecting any three balls would be,
\(
\begin{aligned}
&\Rightarrow{ }^9 C_3=\frac{9 !}{3 !(9-3) !} \\
&\Rightarrow{ }^9 C_3=84
\end{aligned}
\)
Step 2: Find the number of ways of selecting 3 balls without black balls.
There would be 6 balls (if we exclude black balls).
The number of ways of selecting any three balls would be,
\(
\begin{gathered}
\Rightarrow{ }^6 C_3=\frac{6 !}{3 !(6-3) !} \\
\Rightarrow{ }^6 C_3=20
\end{gathered}
\)
Step 3: Find the number of ways 3 balls can be drawn from the box if atleast one black ball is to be included in the draw.
Let, there are \(1 n\) I number of ways 3 balls can be drawn from the box if atleast one black ball is to be included in the draw.
\(
\begin{aligned}
&\Rightarrow n={ }^9 C_3-{ }^6 C_3 \\
&\Rightarrow n=84-20 \\
&\Rightarrow n=64
\end{aligned}
\)

Hint

We have been given 12 points in a plane and in which 5 are collinear.
As we understand in hint if we have given \(\mathrm{n}\) points and we have to make a straight line joining any two of them then the number of ways will be \({ }^n C_2\).

That means if we have 12 points and none of them is collinear then on joining any two of them the number of straight lines will be \({ }^{12} C_2\).

And we also know collinear points means points lying in the same line that means on joining collinear points we will get one and only one line and hence,
We have given 5 collinear points so on joining these 5 points we will get only one line.
Therefore, from the total number of straight lines we have to subtract \({ }^5 C_2\) and add 1 in place of \({ }^5 C_2\) because only one line will form from these 5 points.
Total number of straight lines formed will be
\(
{ }^{12} C_2-{ }^5 C_2+1=57 .
\)
Here \({ }^{12} C_2[latex] is for the number of straight lines formed if no collinear point is present but we subtract [latex]{ }^5 C_2\) because there are 5 collinear points given and we add 1 because on joining collinear points only one line will be formed.

Hint

Each letter can be posted in any one of the five letter boxes. So, the total number of ways of posting three letters \(=5 \times 5 \times 5=125\)

Hint

Arrangement of \(n\) things, taken \(r\) at a time in which \(m\) things occur together. First we select \((r-m)\) objects from \((n-m)\) objects in \({ }^{n-m} C_{r-m}\) ways.
Now we consider these \(m\) things as 1 group.
Number of objects excluding these \(m\) objects \(=(r-m)\)
Now, first we have to arrange \((r-m+1)\) objects.
Number of arrangements \(=(r-m+1)\) !
Also \(m\) objects which we considered as 1 group, can be arranged in \(m\) ! ways.
\(\therefore\) Required number of arrangements \(={ }^{n-m} C_{r-m} \times(r-m+1) ! \times m\) !

Hint

The first stall can be filled in 3 ways, the second in 3 ways, and so on.
Similarly, the 12 th stall can be filled in 3 ways.
Thus the ways of loading the steamer is \(3 \times 3 \times 3 \times 3 \ldots . . .3(12\) times \()=3^{12}\)

Hint

Each of the things can be taken or left out i.e. each gives two ways. So the combinations of \(n\) things will be
\(2 \times 2 \times 2 \ldots \ldots . \mathrm{n}\) factors \(=2^{\mathrm{n}}\) But this includes the case when all have been left out. So the number of combinations is \(2^{\mathrm{n}}-1\).

Hint

There is no restriction about the number of balls (identical) to be drawn except that there will be at least one red ball. Hence we can have 1 or 2 or 3 or 4 red balls in the selection i.e. there are 4 ways of selection of red balls. For black balls there is no restriction. We may select 1 or 2 or 3 or 4 or 5 or no blackball i.e. there are 6 ways of selection of black balls Hence the total number of selection is \(4 \times 6=24\).

Hint

Let two sides of the table be \(A\) and \(B\) each having 9 seats.
Let on side \(A\), four particular guests and on side \(B\), three particular guests be seated.
Now for side \(A\), five more guests can be selected from remaining eleven guests in \({ }^{11} C_5\) ways.
Also on each side nine guests can be arranged in 9! ways.
So total number of ways of arrangements \(={ }^{11} C_5 \times 9 ! \times 9 !=\frac{11 !}{5 ! 6 !}(9 !)(9 !)\)

Hint

A candidate can attempt 5 questions from group I and 2 from group II or 4 from group I and 3 from group II or 3 from group I and 4 from group II or 2 from group I and 5 from group II.
This can be done in
\(
\begin{aligned}
&{ }^6 \mathrm{C}_5 \times{ }^6 \mathrm{C}_2+{ }^6 \mathrm{C}_4 \times{ }^6 \mathrm{C}_3+{ }^6 \mathrm{C}_3 \times{ }^6 \mathrm{C}_4+{ }^6 \mathrm{C}_2 \times{ }^6 \mathrm{C}_5 \\
&=6 \times 15+15 \times 20+20 \times 15+15 \times 6 \\
&=90+300+300+90=780
\end{aligned}
\)

Hint

We can select 3 scheduled caste candidates out of 5 in \({ }^5 C_3\) ways.
And we can select 9 other candidates out of 22 in \({ }^{22} C_9\) ways.
\(\therefore\) Total number of selections \(={ }^5 C_3 \times{ }^{22} C_9\)

Hint

We have 3 books of Mathematics, 4 of Physics and 5 on English
(a) One book of each subject \(={ }^3 C_1 \times{ }^4 C_1 \times{ }^5 C_1\)
\(
\begin{aligned}
&=3 \times 4 \times 5 \\
&=60
\end{aligned}
\)
(b) Atleast one book of each subject \(=\left(2^3-1\right) \times\left(2^4-1\right) \times\left(2^5-1\right)\)
\(
\begin{aligned}
&==7 \times 15 \times 31 \\
&=3255
\end{aligned}
\)
(c) Atleast one book of English \(=\left(2^5-1\right) \times 2^7\)
\(
\begin{aligned}
&=31 \times 128 \\
&=3986
\end{aligned}
\)

Hint

(a) Total number of arrangement when boys and girls alternate: \(=(5 !)^2+(5 !)^2\)
(b) No two girls sit together: = 5! 6!
(c) All the girls sit together = 2! 5! 5!
(d) All the girls sit never together \(=10 !-5 ! 6\) !

Hint

(a) We have to select 2 professor out of 10 and 3 lecturers out of 20
\(\therefore\) Number of ways of selection \(={ }^{10} C_2 \times{ }^{20} C_3\)
(b) When a particular professor is included taken the number of ways \(={ }^{10-1} C_1 \times{ }^{20} C_3\) \(={ }^9 C_1 \times{ }^{20} C_3\)
(c) When a particular lecturer is included then the number of ways \(=10 \mathrm{C}_2 \times{ }^{19} \mathrm{C}_2\)
(d) When a particular lecturer is excluded, then the number of ways \(={ }^{10} C_2 \times{ }^{19} C_3\)

Hint

(a) Total of 4 digit number formed with \(1,2,3,4,5,6,7\)
\(
\begin{aligned}
&={ }^7 P_4 \\
&=\frac{7 \times 6 \times 5 \times 4 \times 3 !}{3 !} \\
&=840
\end{aligned}
\)
(b) When a number is divisible by 2
\(
\begin{aligned}
&=4 \times 5 \times 6 \times 3 \\
&=360
\end{aligned}
\)
(c) Total numbers which are divisible by \(25=40\)
(d) Total numbers which are divisible by 4 (last two digits is divisible by 4 ) \(=200\)

Hint

(a) 4 letters are used at a time \(={ }^6 \mathrm{P}_4\)
\(
\begin{aligned}
&=\frac{6 !}{2 !} \\
&=360
\end{aligned}
\)
(b) All letters are used at a time \(={ }^6 p_6\)
\(=6 !\)
\(=720\)
(c) All letters are used but first letter is vowel \(=2 \times 5\) !
\(=2 \times 120\)
\(=240\)