QBManager

Permutation and Combination

119276 If ${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + 3 {(^{30} C_{3})}^{2} + \dots \dots .$ . $\dots . + 30 {(^{30} C_{30})}^{2} = \frac{α 60!}{30! 30!}$ then $α$ is equal to

1 60

2 10

3 15

4 30

Explanation:

C
${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + 3 {(^{30} C_{3})}^{2} + \dots \dots .$
$+ 30 {(^{30} C_{30})}^{2} = \frac{α 60!}{30! 30!}$
$\Rightarrow$ As we know that
${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + \dots \dots .30 {(^{30} C_{30})}^{2} = \sum_{r = 0}^{30} r {(^{30} C_{r})}^{2}$
$= \sum_{r = 0}^{30} r \cdot^{30} C_{r} \cdot^{30} C_{r}$
$= \sum r \frac{30}{r} \cdot^{29} C_{r - 1} \cdot^{30} C_{r} [^{n} C_{r} = \frac{n}{r} \cdot^{n - 1} C_{r - 1}]$
$= 30 \sum_{r = 1}^{29} C_{r - 1} \cdot^{30} C_{r}$
$= 30 \sum^{29} C_{r - 1} \cdot^{30} C_{30 - r}$
$= 30 \times^{59} C_{29}$
$= 30 \times \frac{59!}{29! \times 30!} \times \frac{60}{30} \times \frac{30}{60}$
$Comparing RHS,$
$15 \times \frac{60!}{(30!) \times (30!)} = \frac{α 60!}{30! 30!}$
Comparing RHS,
Hence, $α = 15$

JEE Main-24.01.2023

Permutation and Combination

119277 If $^{n} p_{4} = 20 \times^{n} p_{2}$ . Then the value of $n$ is

1 18

2 13

3 7

4 4

Explanation:

C We have $^{n} p_{4} = 20 \times^{n} p_{2}$
$\frac{n!}{(n - 4)!} = 20 \times \frac{n!}{(n - 2)!}$
$\frac{1}{(n - 4)!} = 20 \times \frac{1}{(n - 2) (n - 3) (n - 4)!}$
$(n - 2) (n - 3) = 20$
$n^{2} - 3 n - 2 n + 6 = 20$
$n^{2} - 5 n - 14 = 0$
$(n - 7) (n + 2) = 0$
$n = 7$

AMU-2017

Permutation and Combination

119279 The number of words which can be made out of the letters of the word 'MOBILE' when consonants occupy odd places is

1 20

2 36

3 30

4 720

Explanation:

B Given that,
$Total word in MOBILE = 6!$
$Number of consonants = M, B, L = 3$
$Number of vowel = O,I,E = 3$
original image
At odd places 3 consonants take place in 3! ways. and remaining places 3 vowel can take place in 3 ! ways. thus required words $= 3! \times 3! = 6 \times 6 = 36$

APEAPCET- 23.08.2021

Permutation and Combination

119280 If ' $n$ ' is a positive integer, then ${2.4}^{2 n + 1} + 3^{3 n + 1}$ is divisible by

1 2

2 9

3 11

4 27

Explanation:

C : Given that,
$n$ is a positive integer
If $n = 1$
${2.4}^{2 (1) + 1} + 3^{3 (1) + 1}$
$= {2.4}^{3} + 3^{4}$
$= 2.64 + 3^{4}$
$= 2.64 + 81 = 128 + 81 = 209$
Hence, 209 is the divisible by 11 .

APEAPCET- 23.08.2021

Permutation and Combination

119281 $\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots \dots + n \frac{C_{n}}{C_{n - 1}} =$

1

\frac{n (n - 1)}{2}

2

\frac{n (n + 1)}{2}

3

\frac{n (n + 1) (n + 1)}{2}

4 None of these

Explanation:

B
Using formula $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
Then,
$\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots . . + n \frac{C_{n}}{C_{n - 1}}$
$= n + 2 (\frac{n - 1}{2}) + 3 (\frac{n - 2}{3}) + 4 (\frac{n - 3}{4}) + \dots n \cdot \frac{1}{n}$
$= n + (n - 1) + (n - 2) + (n - 3) + \dots . + 1$
$= 1 + 2 + \dots + (n - 2) + (n + 1) + n$
$= \frac{n (n + 1)}{2}$

AMU-2014

Permutation and Combination

119276 If ${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + 3 {(^{30} C_{3})}^{2} + \dots \dots .$ . $\dots . + 30 {(^{30} C_{30})}^{2} = \frac{α 60!}{30! 30!}$ then $α$ is equal to

1 60

2 10

3 15

4 30

Explanation:

C
${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + 3 {(^{30} C_{3})}^{2} + \dots \dots .$
$+ 30 {(^{30} C_{30})}^{2} = \frac{α 60!}{30! 30!}$
$\Rightarrow$ As we know that
${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + \dots \dots .30 {(^{30} C_{30})}^{2} = \sum_{r = 0}^{30} r {(^{30} C_{r})}^{2}$
$= \sum_{r = 0}^{30} r \cdot^{30} C_{r} \cdot^{30} C_{r}$
$= \sum r \frac{30}{r} \cdot^{29} C_{r - 1} \cdot^{30} C_{r} [^{n} C_{r} = \frac{n}{r} \cdot^{n - 1} C_{r - 1}]$
$= 30 \sum_{r = 1}^{29} C_{r - 1} \cdot^{30} C_{r}$
$= 30 \sum^{29} C_{r - 1} \cdot^{30} C_{30 - r}$
$= 30 \times^{59} C_{29}$
$= 30 \times \frac{59!}{29! \times 30!} \times \frac{60}{30} \times \frac{30}{60}$
$Comparing RHS,$
$15 \times \frac{60!}{(30!) \times (30!)} = \frac{α 60!}{30! 30!}$
Comparing RHS,
Hence, $α = 15$

JEE Main-24.01.2023

Permutation and Combination

119277 If $^{n} p_{4} = 20 \times^{n} p_{2}$ . Then the value of $n$ is

1 18

2 13

3 7

4 4

Explanation:

C We have $^{n} p_{4} = 20 \times^{n} p_{2}$
$\frac{n!}{(n - 4)!} = 20 \times \frac{n!}{(n - 2)!}$
$\frac{1}{(n - 4)!} = 20 \times \frac{1}{(n - 2) (n - 3) (n - 4)!}$
$(n - 2) (n - 3) = 20$
$n^{2} - 3 n - 2 n + 6 = 20$
$n^{2} - 5 n - 14 = 0$
$(n - 7) (n + 2) = 0$
$n = 7$

AMU-2017

Permutation and Combination

119279 The number of words which can be made out of the letters of the word 'MOBILE' when consonants occupy odd places is

1 20

2 36

3 30

4 720

Explanation:

B Given that,
$Total word in MOBILE = 6!$
$Number of consonants = M, B, L = 3$
$Number of vowel = O,I,E = 3$
original image
At odd places 3 consonants take place in 3! ways. and remaining places 3 vowel can take place in 3 ! ways. thus required words $= 3! \times 3! = 6 \times 6 = 36$

APEAPCET- 23.08.2021

Permutation and Combination

119280 If ' $n$ ' is a positive integer, then ${2.4}^{2 n + 1} + 3^{3 n + 1}$ is divisible by

1 2

2 9

3 11

4 27

Explanation:

C : Given that,
$n$ is a positive integer
If $n = 1$
${2.4}^{2 (1) + 1} + 3^{3 (1) + 1}$
$= {2.4}^{3} + 3^{4}$
$= 2.64 + 3^{4}$
$= 2.64 + 81 = 128 + 81 = 209$
Hence, 209 is the divisible by 11 .

APEAPCET- 23.08.2021

Permutation and Combination

119281 $\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots \dots + n \frac{C_{n}}{C_{n - 1}} =$

1

\frac{n (n - 1)}{2}

2

\frac{n (n + 1)}{2}

3

\frac{n (n + 1) (n + 1)}{2}

4 None of these

Explanation:

B
Using formula $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
Then,
$\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots . . + n \frac{C_{n}}{C_{n - 1}}$
$= n + 2 (\frac{n - 1}{2}) + 3 (\frac{n - 2}{3}) + 4 (\frac{n - 3}{4}) + \dots n \cdot \frac{1}{n}$
$= n + (n - 1) + (n - 2) + (n - 3) + \dots . + 1$
$= 1 + 2 + \dots + (n - 2) + (n + 1) + n$
$= \frac{n (n + 1)}{2}$

AMU-2014

Permutation and Combination

119276 If ${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + 3 {(^{30} C_{3})}^{2} + \dots \dots .$ . $\dots . + 30 {(^{30} C_{30})}^{2} = \frac{α 60!}{30! 30!}$ then $α$ is equal to

1 60

2 10

3 15

4 30

Explanation:

C
${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + 3 {(^{30} C_{3})}^{2} + \dots \dots .$
$+ 30 {(^{30} C_{30})}^{2} = \frac{α 60!}{30! 30!}$
$\Rightarrow$ As we know that
${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + \dots \dots .30 {(^{30} C_{30})}^{2} = \sum_{r = 0}^{30} r {(^{30} C_{r})}^{2}$
$= \sum_{r = 0}^{30} r \cdot^{30} C_{r} \cdot^{30} C_{r}$
$= \sum r \frac{30}{r} \cdot^{29} C_{r - 1} \cdot^{30} C_{r} [^{n} C_{r} = \frac{n}{r} \cdot^{n - 1} C_{r - 1}]$
$= 30 \sum_{r = 1}^{29} C_{r - 1} \cdot^{30} C_{r}$
$= 30 \sum^{29} C_{r - 1} \cdot^{30} C_{30 - r}$
$= 30 \times^{59} C_{29}$
$= 30 \times \frac{59!}{29! \times 30!} \times \frac{60}{30} \times \frac{30}{60}$
$Comparing RHS,$
$15 \times \frac{60!}{(30!) \times (30!)} = \frac{α 60!}{30! 30!}$
Comparing RHS,
Hence, $α = 15$

JEE Main-24.01.2023

Permutation and Combination

119277 If $^{n} p_{4} = 20 \times^{n} p_{2}$ . Then the value of $n$ is

1 18

2 13

3 7

4 4

Explanation:

C We have $^{n} p_{4} = 20 \times^{n} p_{2}$
$\frac{n!}{(n - 4)!} = 20 \times \frac{n!}{(n - 2)!}$
$\frac{1}{(n - 4)!} = 20 \times \frac{1}{(n - 2) (n - 3) (n - 4)!}$
$(n - 2) (n - 3) = 20$
$n^{2} - 3 n - 2 n + 6 = 20$
$n^{2} - 5 n - 14 = 0$
$(n - 7) (n + 2) = 0$
$n = 7$

AMU-2017

Permutation and Combination

119279 The number of words which can be made out of the letters of the word 'MOBILE' when consonants occupy odd places is

1 20

2 36

3 30

4 720

Explanation:

B Given that,
$Total word in MOBILE = 6!$
$Number of consonants = M, B, L = 3$
$Number of vowel = O,I,E = 3$
original image
At odd places 3 consonants take place in 3! ways. and remaining places 3 vowel can take place in 3 ! ways. thus required words $= 3! \times 3! = 6 \times 6 = 36$

APEAPCET- 23.08.2021

Permutation and Combination

119280 If ' $n$ ' is a positive integer, then ${2.4}^{2 n + 1} + 3^{3 n + 1}$ is divisible by

1 2

2 9

3 11

4 27

Explanation:

C : Given that,
$n$ is a positive integer
If $n = 1$
${2.4}^{2 (1) + 1} + 3^{3 (1) + 1}$
$= {2.4}^{3} + 3^{4}$
$= 2.64 + 3^{4}$
$= 2.64 + 81 = 128 + 81 = 209$
Hence, 209 is the divisible by 11 .

APEAPCET- 23.08.2021

Permutation and Combination

119281 $\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots \dots + n \frac{C_{n}}{C_{n - 1}} =$

1

\frac{n (n - 1)}{2}

2

\frac{n (n + 1)}{2}

3

\frac{n (n + 1) (n + 1)}{2}

4 None of these

Explanation:

B
Using formula $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
Then,
$\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots . . + n \frac{C_{n}}{C_{n - 1}}$
$= n + 2 (\frac{n - 1}{2}) + 3 (\frac{n - 2}{3}) + 4 (\frac{n - 3}{4}) + \dots n \cdot \frac{1}{n}$
$= n + (n - 1) + (n - 2) + (n - 3) + \dots . + 1$
$= 1 + 2 + \dots + (n - 2) + (n + 1) + n$
$= \frac{n (n + 1)}{2}$

AMU-2014

Permutation and Combination

119276 If ${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + 3 {(^{30} C_{3})}^{2} + \dots \dots .$ . $\dots . + 30 {(^{30} C_{30})}^{2} = \frac{α 60!}{30! 30!}$ then $α$ is equal to

1 60

2 10

3 15

4 30

Explanation:

C
${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + 3 {(^{30} C_{3})}^{2} + \dots \dots .$
$+ 30 {(^{30} C_{30})}^{2} = \frac{α 60!}{30! 30!}$
$\Rightarrow$ As we know that
${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + \dots \dots .30 {(^{30} C_{30})}^{2} = \sum_{r = 0}^{30} r {(^{30} C_{r})}^{2}$
$= \sum_{r = 0}^{30} r \cdot^{30} C_{r} \cdot^{30} C_{r}$
$= \sum r \frac{30}{r} \cdot^{29} C_{r - 1} \cdot^{30} C_{r} [^{n} C_{r} = \frac{n}{r} \cdot^{n - 1} C_{r - 1}]$
$= 30 \sum_{r = 1}^{29} C_{r - 1} \cdot^{30} C_{r}$
$= 30 \sum^{29} C_{r - 1} \cdot^{30} C_{30 - r}$
$= 30 \times^{59} C_{29}$
$= 30 \times \frac{59!}{29! \times 30!} \times \frac{60}{30} \times \frac{30}{60}$
$Comparing RHS,$
$15 \times \frac{60!}{(30!) \times (30!)} = \frac{α 60!}{30! 30!}$
Comparing RHS,
Hence, $α = 15$

JEE Main-24.01.2023

Permutation and Combination

119277 If $^{n} p_{4} = 20 \times^{n} p_{2}$ . Then the value of $n$ is

1 18

2 13

3 7

4 4

Explanation:

C We have $^{n} p_{4} = 20 \times^{n} p_{2}$
$\frac{n!}{(n - 4)!} = 20 \times \frac{n!}{(n - 2)!}$
$\frac{1}{(n - 4)!} = 20 \times \frac{1}{(n - 2) (n - 3) (n - 4)!}$
$(n - 2) (n - 3) = 20$
$n^{2} - 3 n - 2 n + 6 = 20$
$n^{2} - 5 n - 14 = 0$
$(n - 7) (n + 2) = 0$
$n = 7$

AMU-2017

Permutation and Combination

119279 The number of words which can be made out of the letters of the word 'MOBILE' when consonants occupy odd places is

1 20

2 36

3 30

4 720

Explanation:

B Given that,
$Total word in MOBILE = 6!$
$Number of consonants = M, B, L = 3$
$Number of vowel = O,I,E = 3$
original image
At odd places 3 consonants take place in 3! ways. and remaining places 3 vowel can take place in 3 ! ways. thus required words $= 3! \times 3! = 6 \times 6 = 36$

APEAPCET- 23.08.2021

Permutation and Combination

119280 If ' $n$ ' is a positive integer, then ${2.4}^{2 n + 1} + 3^{3 n + 1}$ is divisible by

1 2

2 9

3 11

4 27

Explanation:

C : Given that,
$n$ is a positive integer
If $n = 1$
${2.4}^{2 (1) + 1} + 3^{3 (1) + 1}$
$= {2.4}^{3} + 3^{4}$
$= 2.64 + 3^{4}$
$= 2.64 + 81 = 128 + 81 = 209$
Hence, 209 is the divisible by 11 .

APEAPCET- 23.08.2021

Permutation and Combination

119281 $\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots \dots + n \frac{C_{n}}{C_{n - 1}} =$

1

\frac{n (n - 1)}{2}

2

\frac{n (n + 1)}{2}

3

\frac{n (n + 1) (n + 1)}{2}

4 None of these

Explanation:

B
Using formula $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
Then,
$\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots . . + n \frac{C_{n}}{C_{n - 1}}$
$= n + 2 (\frac{n - 1}{2}) + 3 (\frac{n - 2}{3}) + 4 (\frac{n - 3}{4}) + \dots n \cdot \frac{1}{n}$
$= n + (n - 1) + (n - 2) + (n - 3) + \dots . + 1$
$= 1 + 2 + \dots + (n - 2) + (n + 1) + n$
$= \frac{n (n + 1)}{2}$

AMU-2014

Permutation and Combination

119276 If ${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + 3 {(^{30} C_{3})}^{2} + \dots \dots .$ . $\dots . + 30 {(^{30} C_{30})}^{2} = \frac{α 60!}{30! 30!}$ then $α$ is equal to

1 60

2 10

3 15

4 30

Explanation:

C
${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + 3 {(^{30} C_{3})}^{2} + \dots \dots .$
$+ 30 {(^{30} C_{30})}^{2} = \frac{α 60!}{30! 30!}$
$\Rightarrow$ As we know that
${(^{30} C_{1})}^{2} + 2 {(^{30} C_{2})}^{2} + \dots \dots .30 {(^{30} C_{30})}^{2} = \sum_{r = 0}^{30} r {(^{30} C_{r})}^{2}$
$= \sum_{r = 0}^{30} r \cdot^{30} C_{r} \cdot^{30} C_{r}$
$= \sum r \frac{30}{r} \cdot^{29} C_{r - 1} \cdot^{30} C_{r} [^{n} C_{r} = \frac{n}{r} \cdot^{n - 1} C_{r - 1}]$
$= 30 \sum_{r = 1}^{29} C_{r - 1} \cdot^{30} C_{r}$
$= 30 \sum^{29} C_{r - 1} \cdot^{30} C_{30 - r}$
$= 30 \times^{59} C_{29}$
$= 30 \times \frac{59!}{29! \times 30!} \times \frac{60}{30} \times \frac{30}{60}$
$Comparing RHS,$
$15 \times \frac{60!}{(30!) \times (30!)} = \frac{α 60!}{30! 30!}$
Comparing RHS,
Hence, $α = 15$

JEE Main-24.01.2023

Permutation and Combination

119277 If $^{n} p_{4} = 20 \times^{n} p_{2}$ . Then the value of $n$ is

1 18

2 13

3 7

4 4

Explanation:

C We have $^{n} p_{4} = 20 \times^{n} p_{2}$
$\frac{n!}{(n - 4)!} = 20 \times \frac{n!}{(n - 2)!}$
$\frac{1}{(n - 4)!} = 20 \times \frac{1}{(n - 2) (n - 3) (n - 4)!}$
$(n - 2) (n - 3) = 20$
$n^{2} - 3 n - 2 n + 6 = 20$
$n^{2} - 5 n - 14 = 0$
$(n - 7) (n + 2) = 0$
$n = 7$

AMU-2017

Permutation and Combination

119279 The number of words which can be made out of the letters of the word 'MOBILE' when consonants occupy odd places is

1 20

2 36

3 30

4 720

Explanation:

B Given that,
$Total word in MOBILE = 6!$
$Number of consonants = M, B, L = 3$
$Number of vowel = O,I,E = 3$
original image
At odd places 3 consonants take place in 3! ways. and remaining places 3 vowel can take place in 3 ! ways. thus required words $= 3! \times 3! = 6 \times 6 = 36$

APEAPCET- 23.08.2021

Permutation and Combination

119280 If ' $n$ ' is a positive integer, then ${2.4}^{2 n + 1} + 3^{3 n + 1}$ is divisible by

1 2

2 9

3 11

4 27

Explanation:

C : Given that,
$n$ is a positive integer
If $n = 1$
${2.4}^{2 (1) + 1} + 3^{3 (1) + 1}$
$= {2.4}^{3} + 3^{4}$
$= 2.64 + 3^{4}$
$= 2.64 + 81 = 128 + 81 = 209$
Hence, 209 is the divisible by 11 .

APEAPCET- 23.08.2021

Permutation and Combination

119281 $\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots \dots + n \frac{C_{n}}{C_{n - 1}} =$

1

\frac{n (n - 1)}{2}

2

\frac{n (n + 1)}{2}

3

\frac{n (n + 1) (n + 1)}{2}

4 None of these

Explanation:

B
Using formula $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
Then,
$\frac{C_{1}}{C_{0}} + 2 \frac{C_{2}}{C_{1}} + 3 \frac{C_{3}}{C_{2}} + \dots . . + n \frac{C_{n}}{C_{n - 1}}$
$= n + 2 (\frac{n - 1}{2}) + 3 (\frac{n - 2}{3}) + 4 (\frac{n - 3}{4}) + \dots n \cdot \frac{1}{n}$
$= n + (n - 1) + (n - 2) + (n - 3) + \dots . + 1$
$= 1 + 2 + \dots + (n - 2) + (n + 1) + n$
$= \frac{n (n + 1)}{2}$

AMU-2014