Question

Find n such that P(n,5) = 42 P( n,3), n>4​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:P(n,5) = 42P(n,3)$

$\large\underline{\sf{To\:Find - }}$

The value of n

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:P(n,5) = 42P(n,3)$

We know,

$\boxed{ \rm{ P(n,r) = \frac{n!}{(n - r)!} \: where \: n \geqslant r}}$

Using this, we get

$\rm :\longmapsto\:\dfrac{n!}{(n - 5)!} = 42 \times \dfrac{n!}{(n - 3)!}$

$\rm :\longmapsto\:\dfrac{1}{(n - 5)!} = 42 \times \dfrac{1}{(n - 3)(n - 4)(n - 5)!}$

$\rm :\longmapsto\:1 = 42 \times \dfrac{1}{(n - 3)(n - 4)}$

$\rm :\longmapsto\:(n - 3)(n - 4) = 42$

$\rm :\longmapsto\:(n - 3)(n - 4) = 7 \times 6$

On comparing, we get

$\rm :\longmapsto\:n - 3 = 7$

$\bf\implies \:n = 10$

Verification :-

Consider,

$\rm :\longmapsto\:P(n,5)$

$\rm \: = \: \: P(10,5)$

$\rm \: = \: \: \dfrac{10!}{(10 - 5)!}$

$\rm \: = \: \: \dfrac{10!}{5!}$

Now, Consider,

$\rm :\longmapsto\:42 \: P(n,3)$

$\rm \: = \: \: 42 \: P(10,3)$

$\rm \: = \: \: 42 \times \dfrac{10!}{(10 - 3)!}$

$\rm \: = \: \: 42 \times \dfrac{10!}{7!}$

$\rm \: = \: \: 42 \times \dfrac{10!}{7 \times 6 \times 5!}$

$\rm \: = \: \: \dfrac{10!}{5!}$

Hence,

$\rm :\longmapsto\:P(n,5) = 42P(n,3)$

Hence, Verified