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JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Permutations and Combinations

If

^{2 n + 1} P_{n - 1} :^{2 n - 1} P_{n} = 11 : 21

, then

n^{2} + n + 15

is equal to ______

Answer: 45

Solution

Expressing the permutations as factorials,

\frac{^{2 n + 1} P _{n - 1}}{^{2 n - 1} P _{n}} = \frac{( 2 n + 1 )! / ( n + 2 )!}{( 2 n - 1 )! / ( n - 1 )!} = \frac{( 2 n + 1 ) ( 2 n )}{( n + 2 ) ( n + 1 ) n} = \frac{2 ( 2 n + 1 )}{( n + 2 ) ( n + 1 )} .

Setting this equal to

\frac{11}{21}

gives

42 (2 n + 1) = 11 (n + 2) (n + 1)

, i.e.

11 n^{2} - 51 n - 20 = 0

, whose positive root is

n = 5.

Therefore

n^{2} + n + 15 = 25 + 5 + 15 = 45.

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Integer answer

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