QBManager

Permutation and Combination

119311 If $5^{97}$ is divided by 52 , the remainder obtained is

1 3

2 5

3 4

4 0

5 1

Explanation:

B
$5^{97} = 5^{96} \times 5 = {(5^{4})}^{24} \times 5$
$= (625)^{24} \times 5$
$∴ \frac{5^{97}}{52} = \frac{(625)^{24} \times 5}{52}$
If we divide, 625 by 52 ,
Then remainder is 1
$∴$ remainder when $(625)^{24} \times 5$ divided by 52 is 5 .

Kerala CEE-2018

Permutation and Combination

119312 The number of diagonals in a hexagon is

1 8

2 9

3 10

4 11

5 12

Explanation:

B Given
Side of Hexagon, $n = 6$
$Number of diagonals = \frac{n (n - 3)}{2}$
$= \frac{6 (6 - 3)}{2}$
$= 3 \times 3 = 9$

Kerala CEE-2017

Permutation and Combination

119313 The sum
$S = 1 / 9! + 1 / 3! 7! + 1 / 5! 5! + 1 / 7! 3! + 1 / 9$ ! is equal to

1

2^{10} / 8

!

2

2^{9} / 10

!

3

2^{7} / 10

!

4

2^{6} / 10

!

5

2^{5} / 8

!

Explanation:

B Sum
$S = \frac{1}{9!} + \frac{1}{3! 7!} + \frac{1}{5! 5!} + \frac{1}{7! 3!} + \frac{1}{9!}$
$S = \frac{1}{10!} [\frac{10!}{9!} + \frac{10!}{3! 7!} + \frac{10!}{5! 5!} + \frac{10!}{7! 3!} + \frac{10!}{9!}]$
$= \frac{1}{10!} [^{10} C_{1} +^{10} C_{3} +^{10} C_{5} +^{10} C_{7} +^{10} C_{9}]$
$= \frac{1}{10!} (2^{10 - 1})$
$= \frac{2^{9}}{10!}$

Kerala CEE-2017

Permutation and Combination

119316 The value of $\frac{1}{2!} + \frac{2}{3!} + \dots + \frac{999}{1000!}$ is equal to

1

\frac{1000! - 1}{1000!}

2

\frac{1000! + 1}{1000!}

3

\frac{999! - 1}{999!}

4

\frac{999! + 1}{999!}

5

\frac{1000! - 999!}{1000!}

Explanation:

A Given
$= \frac{1}{2!} + \frac{2}{3!} + \dots \dots + \frac{999}{1000!}$
$= \frac{2 - 1}{2!} + \frac{3 - 1}{3!} + \dots \dots \dots + \frac{1000 - 1}{1000!}$
$= \frac{2}{2!} - \frac{1}{2!} + \frac{3}{3!} - \frac{1}{3!} + \dots \dots \dots + \frac{1000}{1000!} - \frac{1}{1000!}$
$= 1 - \frac{1}{2!} + \frac{1}{2!} - \frac{1}{3!} + \dots \dots + \frac{1}{999} - \frac{1}{1000}$
$∵ \frac{n}{n!} = \frac{1}{(n - 1)!}]$
$= 1 - \frac{1}{1000!} = \frac{1000! - 1}{1000!}$

Kerala CEE-2009

Permutation and Combination

119314 If $n$ is any positive integer, the $\frac{1}{2^{n}} (^{2 n} P_{n})$ is equal to

1

2 \cdot 4 \cdot 6 \dot{} (2 n)

2

1 \cdot 2 \cdot 3 \dot{}

n

3

1 \cdot 3 \cdot 5 \dot{} (2 n - 1)

4

1 \cdot 2 \cdot 3 \dot{} (3 n)

5

2 \cdot 4 \cdot 6 \dot{} (2 n + 2)

Explanation:

C Positive integer
$= \frac{1}{2^{n}} (^{2 n} P_{n}) = \frac{2 n (2 n - 1) (2 n - 2) (2 n - 3) \dots .3 \times 2 \times 1}{2^{n} \times n!}$
$= \frac{2^{n} \times n (n - 1) (n - 2) \dots .3 \times 2 \times 2 \times 1 [1 \times 3 \times 5 (2 n - 1)]}{2^{n} n!}$
$= 1 \times 3 \times 5 \dots \dots (2 n - 1)$

Kerala CEE-2012

Permutation and Combination

119311 If $5^{97}$ is divided by 52 , the remainder obtained is

1 3

2 5

3 4

4 0

5 1

Explanation:

B
$5^{97} = 5^{96} \times 5 = {(5^{4})}^{24} \times 5$
$= (625)^{24} \times 5$
$∴ \frac{5^{97}}{52} = \frac{(625)^{24} \times 5}{52}$
If we divide, 625 by 52 ,
Then remainder is 1
$∴$ remainder when $(625)^{24} \times 5$ divided by 52 is 5 .

Kerala CEE-2018

Permutation and Combination

119312 The number of diagonals in a hexagon is

1 8

2 9

3 10

4 11

5 12

Explanation:

B Given
Side of Hexagon, $n = 6$
$Number of diagonals = \frac{n (n - 3)}{2}$
$= \frac{6 (6 - 3)}{2}$
$= 3 \times 3 = 9$

Kerala CEE-2017

Permutation and Combination

119313 The sum
$S = 1 / 9! + 1 / 3! 7! + 1 / 5! 5! + 1 / 7! 3! + 1 / 9$ ! is equal to

1

2^{10} / 8

!

2

2^{9} / 10

!

3

2^{7} / 10

!

4

2^{6} / 10

!

5

2^{5} / 8

!

Explanation:

B Sum
$S = \frac{1}{9!} + \frac{1}{3! 7!} + \frac{1}{5! 5!} + \frac{1}{7! 3!} + \frac{1}{9!}$
$S = \frac{1}{10!} [\frac{10!}{9!} + \frac{10!}{3! 7!} + \frac{10!}{5! 5!} + \frac{10!}{7! 3!} + \frac{10!}{9!}]$
$= \frac{1}{10!} [^{10} C_{1} +^{10} C_{3} +^{10} C_{5} +^{10} C_{7} +^{10} C_{9}]$
$= \frac{1}{10!} (2^{10 - 1})$
$= \frac{2^{9}}{10!}$

Kerala CEE-2017

Permutation and Combination

119316 The value of $\frac{1}{2!} + \frac{2}{3!} + \dots + \frac{999}{1000!}$ is equal to

1

\frac{1000! - 1}{1000!}

2

\frac{1000! + 1}{1000!}

3

\frac{999! - 1}{999!}

4

\frac{999! + 1}{999!}

5

\frac{1000! - 999!}{1000!}

Explanation:

A Given
$= \frac{1}{2!} + \frac{2}{3!} + \dots \dots + \frac{999}{1000!}$
$= \frac{2 - 1}{2!} + \frac{3 - 1}{3!} + \dots \dots \dots + \frac{1000 - 1}{1000!}$
$= \frac{2}{2!} - \frac{1}{2!} + \frac{3}{3!} - \frac{1}{3!} + \dots \dots \dots + \frac{1000}{1000!} - \frac{1}{1000!}$
$= 1 - \frac{1}{2!} + \frac{1}{2!} - \frac{1}{3!} + \dots \dots + \frac{1}{999} - \frac{1}{1000}$
$∵ \frac{n}{n!} = \frac{1}{(n - 1)!}]$
$= 1 - \frac{1}{1000!} = \frac{1000! - 1}{1000!}$

Kerala CEE-2009

Permutation and Combination

119314 If $n$ is any positive integer, the $\frac{1}{2^{n}} (^{2 n} P_{n})$ is equal to

1

2 \cdot 4 \cdot 6 \dot{} (2 n)

2

1 \cdot 2 \cdot 3 \dot{}

n

3

1 \cdot 3 \cdot 5 \dot{} (2 n - 1)

4

1 \cdot 2 \cdot 3 \dot{} (3 n)

5

2 \cdot 4 \cdot 6 \dot{} (2 n + 2)

Explanation:

C Positive integer
$= \frac{1}{2^{n}} (^{2 n} P_{n}) = \frac{2 n (2 n - 1) (2 n - 2) (2 n - 3) \dots .3 \times 2 \times 1}{2^{n} \times n!}$
$= \frac{2^{n} \times n (n - 1) (n - 2) \dots .3 \times 2 \times 2 \times 1 [1 \times 3 \times 5 (2 n - 1)]}{2^{n} n!}$
$= 1 \times 3 \times 5 \dots \dots (2 n - 1)$

Kerala CEE-2012

Permutation and Combination

119311 If $5^{97}$ is divided by 52 , the remainder obtained is

1 3

2 5

3 4

4 0

5 1

Explanation:

B
$5^{97} = 5^{96} \times 5 = {(5^{4})}^{24} \times 5$
$= (625)^{24} \times 5$
$∴ \frac{5^{97}}{52} = \frac{(625)^{24} \times 5}{52}$
If we divide, 625 by 52 ,
Then remainder is 1
$∴$ remainder when $(625)^{24} \times 5$ divided by 52 is 5 .

Kerala CEE-2018

Permutation and Combination

119312 The number of diagonals in a hexagon is

1 8

2 9

3 10

4 11

5 12

Explanation:

B Given
Side of Hexagon, $n = 6$
$Number of diagonals = \frac{n (n - 3)}{2}$
$= \frac{6 (6 - 3)}{2}$
$= 3 \times 3 = 9$

Kerala CEE-2017

Permutation and Combination

119313 The sum
$S = 1 / 9! + 1 / 3! 7! + 1 / 5! 5! + 1 / 7! 3! + 1 / 9$ ! is equal to

1

2^{10} / 8

!

2

2^{9} / 10

!

3

2^{7} / 10

!

4

2^{6} / 10

!

5

2^{5} / 8

!

Explanation:

B Sum
$S = \frac{1}{9!} + \frac{1}{3! 7!} + \frac{1}{5! 5!} + \frac{1}{7! 3!} + \frac{1}{9!}$
$S = \frac{1}{10!} [\frac{10!}{9!} + \frac{10!}{3! 7!} + \frac{10!}{5! 5!} + \frac{10!}{7! 3!} + \frac{10!}{9!}]$
$= \frac{1}{10!} [^{10} C_{1} +^{10} C_{3} +^{10} C_{5} +^{10} C_{7} +^{10} C_{9}]$
$= \frac{1}{10!} (2^{10 - 1})$
$= \frac{2^{9}}{10!}$

Kerala CEE-2017

Permutation and Combination

119316 The value of $\frac{1}{2!} + \frac{2}{3!} + \dots + \frac{999}{1000!}$ is equal to

1

\frac{1000! - 1}{1000!}

2

\frac{1000! + 1}{1000!}

3

\frac{999! - 1}{999!}

4

\frac{999! + 1}{999!}

5

\frac{1000! - 999!}{1000!}

Explanation:

A Given
$= \frac{1}{2!} + \frac{2}{3!} + \dots \dots + \frac{999}{1000!}$
$= \frac{2 - 1}{2!} + \frac{3 - 1}{3!} + \dots \dots \dots + \frac{1000 - 1}{1000!}$
$= \frac{2}{2!} - \frac{1}{2!} + \frac{3}{3!} - \frac{1}{3!} + \dots \dots \dots + \frac{1000}{1000!} - \frac{1}{1000!}$
$= 1 - \frac{1}{2!} + \frac{1}{2!} - \frac{1}{3!} + \dots \dots + \frac{1}{999} - \frac{1}{1000}$
$∵ \frac{n}{n!} = \frac{1}{(n - 1)!}]$
$= 1 - \frac{1}{1000!} = \frac{1000! - 1}{1000!}$

Kerala CEE-2009

Permutation and Combination

119314 If $n$ is any positive integer, the $\frac{1}{2^{n}} (^{2 n} P_{n})$ is equal to

1

2 \cdot 4 \cdot 6 \dot{} (2 n)

2

1 \cdot 2 \cdot 3 \dot{}

n

3

1 \cdot 3 \cdot 5 \dot{} (2 n - 1)

4

1 \cdot 2 \cdot 3 \dot{} (3 n)

5

2 \cdot 4 \cdot 6 \dot{} (2 n + 2)

Explanation:

C Positive integer
$= \frac{1}{2^{n}} (^{2 n} P_{n}) = \frac{2 n (2 n - 1) (2 n - 2) (2 n - 3) \dots .3 \times 2 \times 1}{2^{n} \times n!}$
$= \frac{2^{n} \times n (n - 1) (n - 2) \dots .3 \times 2 \times 2 \times 1 [1 \times 3 \times 5 (2 n - 1)]}{2^{n} n!}$
$= 1 \times 3 \times 5 \dots \dots (2 n - 1)$

Kerala CEE-2012

Permutation and Combination

119311 If $5^{97}$ is divided by 52 , the remainder obtained is

1 3

2 5

3 4

4 0

5 1

Explanation:

B
$5^{97} = 5^{96} \times 5 = {(5^{4})}^{24} \times 5$
$= (625)^{24} \times 5$
$∴ \frac{5^{97}}{52} = \frac{(625)^{24} \times 5}{52}$
If we divide, 625 by 52 ,
Then remainder is 1
$∴$ remainder when $(625)^{24} \times 5$ divided by 52 is 5 .

Kerala CEE-2018

Permutation and Combination

119312 The number of diagonals in a hexagon is

1 8

2 9

3 10

4 11

5 12

Explanation:

B Given
Side of Hexagon, $n = 6$
$Number of diagonals = \frac{n (n - 3)}{2}$
$= \frac{6 (6 - 3)}{2}$
$= 3 \times 3 = 9$

Kerala CEE-2017

Permutation and Combination

119313 The sum
$S = 1 / 9! + 1 / 3! 7! + 1 / 5! 5! + 1 / 7! 3! + 1 / 9$ ! is equal to

1

2^{10} / 8

!

2

2^{9} / 10

!

3

2^{7} / 10

!

4

2^{6} / 10

!

5

2^{5} / 8

!

Explanation:

B Sum
$S = \frac{1}{9!} + \frac{1}{3! 7!} + \frac{1}{5! 5!} + \frac{1}{7! 3!} + \frac{1}{9!}$
$S = \frac{1}{10!} [\frac{10!}{9!} + \frac{10!}{3! 7!} + \frac{10!}{5! 5!} + \frac{10!}{7! 3!} + \frac{10!}{9!}]$
$= \frac{1}{10!} [^{10} C_{1} +^{10} C_{3} +^{10} C_{5} +^{10} C_{7} +^{10} C_{9}]$
$= \frac{1}{10!} (2^{10 - 1})$
$= \frac{2^{9}}{10!}$

Kerala CEE-2017

Permutation and Combination

119316 The value of $\frac{1}{2!} + \frac{2}{3!} + \dots + \frac{999}{1000!}$ is equal to

1

\frac{1000! - 1}{1000!}

2

\frac{1000! + 1}{1000!}

3

\frac{999! - 1}{999!}

4

\frac{999! + 1}{999!}

5

\frac{1000! - 999!}{1000!}

Explanation:

A Given
$= \frac{1}{2!} + \frac{2}{3!} + \dots \dots + \frac{999}{1000!}$
$= \frac{2 - 1}{2!} + \frac{3 - 1}{3!} + \dots \dots \dots + \frac{1000 - 1}{1000!}$
$= \frac{2}{2!} - \frac{1}{2!} + \frac{3}{3!} - \frac{1}{3!} + \dots \dots \dots + \frac{1000}{1000!} - \frac{1}{1000!}$
$= 1 - \frac{1}{2!} + \frac{1}{2!} - \frac{1}{3!} + \dots \dots + \frac{1}{999} - \frac{1}{1000}$
$∵ \frac{n}{n!} = \frac{1}{(n - 1)!}]$
$= 1 - \frac{1}{1000!} = \frac{1000! - 1}{1000!}$

Kerala CEE-2009

Permutation and Combination

119314 If $n$ is any positive integer, the $\frac{1}{2^{n}} (^{2 n} P_{n})$ is equal to

1

2 \cdot 4 \cdot 6 \dot{} (2 n)

2

1 \cdot 2 \cdot 3 \dot{}

n

3

1 \cdot 3 \cdot 5 \dot{} (2 n - 1)

4

1 \cdot 2 \cdot 3 \dot{} (3 n)

5

2 \cdot 4 \cdot 6 \dot{} (2 n + 2)

Explanation:

C Positive integer
$= \frac{1}{2^{n}} (^{2 n} P_{n}) = \frac{2 n (2 n - 1) (2 n - 2) (2 n - 3) \dots .3 \times 2 \times 1}{2^{n} \times n!}$
$= \frac{2^{n} \times n (n - 1) (n - 2) \dots .3 \times 2 \times 2 \times 1 [1 \times 3 \times 5 (2 n - 1)]}{2^{n} n!}$
$= 1 \times 3 \times 5 \dots \dots (2 n - 1)$

Kerala CEE-2012

Permutation and Combination

119311 If $5^{97}$ is divided by 52 , the remainder obtained is

1 3

2 5

3 4

4 0

5 1

Explanation:

B
$5^{97} = 5^{96} \times 5 = {(5^{4})}^{24} \times 5$
$= (625)^{24} \times 5$
$∴ \frac{5^{97}}{52} = \frac{(625)^{24} \times 5}{52}$
If we divide, 625 by 52 ,
Then remainder is 1
$∴$ remainder when $(625)^{24} \times 5$ divided by 52 is 5 .

Kerala CEE-2018

Permutation and Combination

119312 The number of diagonals in a hexagon is

1 8

2 9

3 10

4 11

5 12

Explanation:

B Given
Side of Hexagon, $n = 6$
$Number of diagonals = \frac{n (n - 3)}{2}$
$= \frac{6 (6 - 3)}{2}$
$= 3 \times 3 = 9$

Kerala CEE-2017

Permutation and Combination

119313 The sum
$S = 1 / 9! + 1 / 3! 7! + 1 / 5! 5! + 1 / 7! 3! + 1 / 9$ ! is equal to

1

2^{10} / 8

!

2

2^{9} / 10

!

3

2^{7} / 10

!

4

2^{6} / 10

!

5

2^{5} / 8

!

Explanation:

B Sum
$S = \frac{1}{9!} + \frac{1}{3! 7!} + \frac{1}{5! 5!} + \frac{1}{7! 3!} + \frac{1}{9!}$
$S = \frac{1}{10!} [\frac{10!}{9!} + \frac{10!}{3! 7!} + \frac{10!}{5! 5!} + \frac{10!}{7! 3!} + \frac{10!}{9!}]$
$= \frac{1}{10!} [^{10} C_{1} +^{10} C_{3} +^{10} C_{5} +^{10} C_{7} +^{10} C_{9}]$
$= \frac{1}{10!} (2^{10 - 1})$
$= \frac{2^{9}}{10!}$

Kerala CEE-2017

Permutation and Combination

119316 The value of $\frac{1}{2!} + \frac{2}{3!} + \dots + \frac{999}{1000!}$ is equal to

1

\frac{1000! - 1}{1000!}

2

\frac{1000! + 1}{1000!}

3

\frac{999! - 1}{999!}

4

\frac{999! + 1}{999!}

5

\frac{1000! - 999!}{1000!}

Explanation:

A Given
$= \frac{1}{2!} + \frac{2}{3!} + \dots \dots + \frac{999}{1000!}$
$= \frac{2 - 1}{2!} + \frac{3 - 1}{3!} + \dots \dots \dots + \frac{1000 - 1}{1000!}$
$= \frac{2}{2!} - \frac{1}{2!} + \frac{3}{3!} - \frac{1}{3!} + \dots \dots \dots + \frac{1000}{1000!} - \frac{1}{1000!}$
$= 1 - \frac{1}{2!} + \frac{1}{2!} - \frac{1}{3!} + \dots \dots + \frac{1}{999} - \frac{1}{1000}$
$∵ \frac{n}{n!} = \frac{1}{(n - 1)!}]$
$= 1 - \frac{1}{1000!} = \frac{1000! - 1}{1000!}$

Kerala CEE-2009

Permutation and Combination

119314 If $n$ is any positive integer, the $\frac{1}{2^{n}} (^{2 n} P_{n})$ is equal to

1

2 \cdot 4 \cdot 6 \dot{} (2 n)

2

1 \cdot 2 \cdot 3 \dot{}

n

3

1 \cdot 3 \cdot 5 \dot{} (2 n - 1)

4

1 \cdot 2 \cdot 3 \dot{} (3 n)

5

2 \cdot 4 \cdot 6 \dot{} (2 n + 2)

Explanation:

C Positive integer
$= \frac{1}{2^{n}} (^{2 n} P_{n}) = \frac{2 n (2 n - 1) (2 n - 2) (2 n - 3) \dots .3 \times 2 \times 1}{2^{n} \times n!}$
$= \frac{2^{n} \times n (n - 1) (n - 2) \dots .3 \times 2 \times 2 \times 1 [1 \times 3 \times 5 (2 n - 1)]}{2^{n} n!}$
$= 1 \times 3 \times 5 \dots \dots (2 n - 1)$

Kerala CEE-2012

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