Quiz Level-2

Hint

Since clockwise and anticlockwise are same in case of ring.
If the anti-clockwise and clockwise sequence and movement are not different then the number of circular permutations of \(n\) distinct items is
\(
=(n-1) ! / 2
\)
Required number of ways \(=(5-1) ! / 2=4 ! / 2=12\)

Hint

By the fundamental property of circular permutation
The count of circular permutations of \(n\) different things are ( \(n-1)\) !

Since two positions are fixed so we have
The required number of ways \((11-2) ! \times 2=9 ! \times 2=725760\)

Hint

The count of circular permutations of n different things are (n-1)!

The number of ways in which 6 men can be arranged at a round table = (6 – 1)!  =5!

Now women can be arranged in 6! ways and the Total Number of ways = 6! × 5!

Hint

If we fix one man round a table then their permutations \(={ }^4 \mathrm{P}_4=24\)
Now if women sit between the two men then their permutations \(={ }^5 \mathrm{P}_5=120\)
So total permutations \(=24 \times 120=2880\)

Hint

Number of keys \(=4\)
Fixing one key we have permutation \(={ }^3 \mathrm{P}_3=6\)
Since the above figures of arrangement are reflections of each other. Therefore permutations \(=1 / 2 \times 6=3\)

Hint

Number of beads \(=6\)
Fixing one bead, we have permutation \(={ }^5 \mathrm{P}_5=120\)
Since above figures of arrangement are reflections of each other, therefore permutations \(=1 / 2 \times 120=60\)

Hint

Boys can sit on the circular table in (4-1)! ways. The girls can be seated in 4 places in 4! ways. So. the number of ways =3!×4!

Hint

Fixing one officer on a particular seat, we have permutations of the remaining 11 officers.
=(12-1)!=11!=39916800

Hint

Number of D.C.O’s \(=9\)
Let \(D_1\) and \(D_2\) be the two D.C.O’s insisting to sit together so consider them one.
If \(D_1 D_2\) sit together then permutations
\(
={ }^9 P_9=362880
\)
If \(D_2 D_1\) sit together then permutations
\(
={ }^9 P_9=362880
\)
So total permutations \(=362880+362880\) \(=725760\)

Alternate:

\(
2 ! \cdot(11-2) !=2 \times 362880=725760
\)

Hint

9 males can be seated on a round table
\(
={ }^8 P_8=40320
\)
And 5 females can be seated on a round table
\(
={ }^4 P_4=24
\)
So permutations of both \(=40320 \times 24\)
\(
=967680 .
\)

Alternate:
\(
=(9-1) ! \times(5-1) !
\)

Hint

PAKPATTAN
Total number of letters \(=9\)
\(\mathrm{P}\) is repeated 2 times
A is repeated 3 times
\(\mathrm{T}\) is repeated 2 times
\(\mathrm{K}\) and \(\mathrm{N}\) come only once.
Required number of permutations =
\(
\begin{aligned}
& =\frac{9 !}{2 ! \times 3 ! \times 2 \backslash \times 1 \times 1 !} \\
& =\frac{362880}{(2)(6)(2)}=15120 .
\end{aligned}
\)

Hint

Total number of digits \(=6\)
Number of 2’s \(=2\)
Number of 3’s =2
Number of 4 ‘s \(=2\)
So number formed by these 6 digits
\(
\begin{aligned}
& =\frac{6 !}{(2 !)(2 !)(2 !)} \\
& =\frac{720}{(2)(2)(2)}=90 .
\end{aligned}
\)

Hint

Total number of digits \(=6\)
Number of 2’s \(=2\)
Number of 3’s =2
Number of 4 ‘s \(=2\)
So number formed by these 6 digits
\(
\begin{aligned}
& =\frac{6 !}{(2 !)(2 !)(2 !)} \\
& =\frac{720}{(2)(2)(2)}=90 .
\end{aligned}
\)
The numbers lie between 400,000 and 430,000 are only of the form \(42^{* * * *}\), where each * can be filled by \(2,3,3,4\).
Here number of digits \(=4\).
Number of 2’s \(=1\)
Number of 3’s \(=2\)
Number of 4’s = 1
\(
\begin{aligned}
\text { So number formed }
& =\frac{4 !}{(1 !)(2 !)(1 !)} \\
& =\frac{24}{2}=12
\end{aligned}
\)

Hint

Total members =11
Members in first committee =3
Members in second committee =4
Members in third committee =2
Members in fourth committee =2
So required number of committees
\(
\begin{aligned}
& =\frac{11 !}{3 ! \cdot 4 ! \cdot 2 ! \cdot 2 !} \\
& =\frac{39916800}{(6)(24)(2)(2)}=69300 .
\end{aligned}
\)

Hint

If \(C\) be the first letter and \(K\) is the last letter then words are of the form \(\mathrm{C} * * * * * * \mathrm{~K}\). where each * can be replaced with A, T, T, A, E, D.
So number of letters \(=6\)
\(\mathrm{A}\) is repeated 2 times
\(\mathrm{T}\) is repeated 2 times
\(E\) and D come only once.
\(
\begin{gathered}
\text { So required permutations }
& =\frac{6 !}{2 ! \times 2 \times 1 \times 1 !}=\frac{720}{4}=180 .
\end{gathered}
\)

Hint

The number greater than 1000000 are of the following forms.
If numbers are of the form \(2 * * * * * *\),
where each \(*\) can be filled with \(0,2,2,3,4,4\)
Then number of digits \(=6\)
2 is repeated 2 times
4 is repeated 2 times
0 and 3 come only once.
So number formed
\(
=\frac{6 !}{2 ! \times 2 \times 1 \times 1 \times 1 !}=\frac{720}{4}=180 \text {. }
\)
Now if numbers are of the form \(3 * * * * *\), where each \(*\) can be filled with \(0,2,2,2,4,4\)
Then number of digits \(=6\)
2 is repeated 3 times
4 is repeated 2 times
0 comes only once.
So number formed
\(
=\frac{6 !}{3 ! \times 2 \times 1 !}=\frac{720}{12}=60 \text {. }
\)
Now if numbers are of the form \(4 * * * * * *\), where each \(*\) can be filled with \(0,2,2,2,3,4\)
Then number of digits \(=6\)
2 is repeated 3 times
0,3 and 4 come only once.
So number formed \(=\frac{6 !}{3 ! \times 1 \times 1 !}\)
\(
=\frac{720}{6}=120 \text {. }
\)
So required numbers greater than 1000000
\(
\begin{aligned}
& =180+60+120 \\
& =360 .
\end{aligned}
\)

Hint

If \(P\) is the first letter then words are of the form \(\mathrm{P} * * * * *\), where five \(*\) can be replace with \(\mathrm{A}, \mathrm{N}, \mathrm{A}, \mathrm{M}, \mathrm{A}\).
So number of letters \(=5\)
\(\mathrm{A}\) is repeated 3 times
\(\mathrm{M}, \mathrm{N}\) appears only once
\(
\begin{aligned}
\text { So required permutations }
& =\frac{5 !}{3 ! \times 1 ! \times 1 !} \\
& =\frac{120}{6}=20 .
\end{aligned}
\)

Hint

A necklace of 16 beads can be arranged in \(15 ! / 2\) ways since a necklace is same when turned upside down. Now it has 5 and 3 similar beads. Hence the total number of distinct necklaces that can be made
\(
=\frac{15 !}{2 \times 5 ! \times 3 !}=\frac{1}{16} \times \frac{16 !}{2 \times 5 ! \times 3 !}
\)
The following solution below shows that number of clockwise circular permutations is equal to the number of anticlockwise circular permutations

Hint

By theory, let us consider 3 beads as one. Hence we have, in effect, 21 beads, ‘n’ = 21.

The number of arrangements = (n − 1)! = 20!

Also, the number of ways in which 3 beads can be arranged between themselves is 3! = 3 x 2 x 1 = 6.

Thus the total number of arrangements = (1/2) x 20! x 3!

Hint

10 boys can be seated in a circle in 9! ways. There are 10 spaces in between the boys, which can be occupied by 5 girls in \({ }^{10} p_5\) ways.
Hence the total number of ways
\(
=9 !{ }^{10} p_5=9 ! 10 ! / 5 !
\)

Hint

In case of circular table the clockwise and anticlockwise arrangements will be treated as different arrangements.

Hence the total number of ways is

\(
{ }^{15} \mathrm{P}_9 / 9
\)

Hint

In the given word, we treat the vowels IEO as 1 letter.
Thus, we have DRCTR (IEO).
This group has 6 letters in which R occurs 2 times and others are all different.

Number of ways of arranging these letters = 6 ! / 2 !

\(
\frac{(6 \times 5 \times 4 \times 3 \times 2 \times 1)}{2}=360
\)

Now, 3 vowels can be arranged among themselves in 3 ! = 6 ways.

Required number of ways = (360 × 6) = 2160.

Hint

In the given word DIGEST, we take the vowels IE as one letter.
Then, we can write it as DGST (IE).
This word has 5 letters which can be arranged among themselves in
5! = (5 × 4 × 3 × 2 × 1) = 120 ways.
The letters of IE can be arranged in 2 ways.
∴ Number of ways of arranging the letters of given word with vowels together
= (120 × 2) = 240 ways.
Number of ways of arranging all the letters of the given word
= 6! = (6 × 5 × 4 × 3 × 2 × 1) = 720 ways.
∴ Number of ways of arrangements so that the vowels are never together
= (720 – 240) = 480.

Hint

There are 6 letters in the given word, out of which there are 3 consonants and 3 vowels.
Let us mark these positions as (1) (2) (3) (4) (5) (6).
Now, 3 vowels can be placed at any of 3 places, marked 1, 3,5 .
Number of these arrangements \(={ }^3 P_3=3!=6\).
Also, 3 consonants can be placed at the remaining 3 places.
Number of these arrangements \(={ }^3 P_3=3!=6\).
Total number of ways \(=(6 \times 6)=36\).

Hint

The given word contains 5 different letters.
Keeping the vowels UE together, we suppose them as 1 letter.
Then, we have to arrange the letters JDG(UE).
Now, 4 letters can be arranged in \(4!=24\) ways.
The vowels (UE) can be arranged among themselves in 2 ways.
\(\therefore\) Required no. of ways \(=(24 \times 2)=48\)

Hint

There are 2 primary teachers.
They can stand in a row in P(2, 2) = 2! = 2 × 1 ways = 2 ways
∴ Two middle teachers.
They can stand in a row in P(2, 2) = 2! = 2 × 1 = 2 ways
There are two secondary teachers.
They can stand in a row in P(2, 2) = 2! = 2 × 1 = 2 ways
These three sets can be arranged themselves in 3! Ways = 3 × 2 × 1 = 6 ways
Hence, the required number of ways
= 2 × 2 × 2 × 6 = 48 ways

Hint

5 students can win the 1st place, leaving 4 who can place second and then 3 for third place. This makes \(5 \cdot 4 \cdot 3=60\) ways that students can place \(1^{st}, 2^{\text {nd }}\), and \(3^{\text {rd }}\)

Hint

The first pentagon can be one of seven different colors, the second can be any one of six, and there are five colors which remain for the final pentagon. \(7 \cdot 6 \cdot 5=210\) ways to color the logo.

Hint

There are 5 letters to choose from for the first spot, leaving 4 for the second, 3 for the third, 2 for the fourth, and there is 1 remaining for the end of the word: \(P(5,5)=5 !=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120\) “words”.

Hint

Notice that there are two A’s. No need to panic, just pretend for a moment that the A’s are different. We’ll call one of them \(A_1\) and the other \(A_2\). \(P(7,7)=7 !=7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=5,040\) “words”.
However, we over-counted. A LGEBRA is the same as \(A_2 L G E B R A_1\), just like \(A_1 A_2 L E G B R\) is the same as \(A_2 A_1\) LEGBR. To eliminate the extra cases, we need to divide by the ways that the A’s can be arranged, which in this case is just \(2 !=2\).
\(
\frac{P(7,7)}{2}=\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2}=2,520 \text { “words”. }
\)