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

1.	Find n such that P(n,5) = 42 P( n,3), n>4
Answer» $\large\underline{\sf{Given- }}$ $\rm :\longmapsto\:P(n,5) = 42P(n,3)$ $\large\underline{\sf{To\:Find - }}$ 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

Answer»

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