Question

find the positive integer n such that
n! /n-2!=240​

Answer 1

SOLUTION

TO DETERMINE

The positive integer n such that

$\displaystyle \sf{ \frac{n!}{(n - 2)!} = 240\: }$

EVALUATION

Here it is given that

$\displaystyle \sf{ \frac{n!}{(n - 2)!} = 240\: }$

$\implies \displaystyle \sf{ \frac{n(n - 1) \times (n - 2)!}{(n - 2)!} = 240\: }$

$\implies \displaystyle \sf{ n(n - 1) = 240\: }$

$\implies \displaystyle \sf{ {n}^{2} - n - 240 = 0\: }$

$\implies \displaystyle \sf{ {n}^{2} - 16n + 15n - 240 = 0\: }$

$\implies \displaystyle \sf{ n(n - 16) + 15(n - 16) = 0\: }$

$\implies \displaystyle \sf{ (n - 16) (n + 15) = 0\: }$

Either n - 16 = 0 or n + 15 = 0

Now n - 16 = 0 gives n = 16

Again n + 15 = 0 gives n = - 15

Since n is a positive integer

So n # - 15

$\therefore \: \: \sf{n = 16}$

FINAL ANSWER

The required value of n = 16

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