Question

find using permutation and combination​

Answer 1

Question :-

Find the value of n using permutation and combination -

$\sf ^nP_4 : ^{n-1}P_3 = 9 : 1$

Answer :-

We know that,

$\implies\sf ^nP_r = \dfrac{n!}{(n-r)!}$

So,

$\implies\sf ^nP_4 : ^{n-1}P_3 = 9 : 1$

$\implies\sf \dfrac{^nP_4}{ ^{n-1}P_3} = \dfrac{9}{1}$

$\implies\sf \dfrac{\dfrac{n!}{(n-4)!}}{\dfrac{(n-1)!}{(n-1-3)!}} = 9$

$\implies\sf \dfrac{\dfrac{n!}{\cancel{(n-4)!}}}{\dfrac{(n-1)!}{\cancel{(n-4)!}}} = 9$

$\implies\sf \dfrac{n!}{(n-1)!} = 9$

$\implies\sf \dfrac{n \times \cancel{(n-1)!}}{\cancel{(n-1)!}} = 9$

$\implies\boxed{\sf n = 9}$

Verification :-

$\sf LHS = \dfrac{^nP_4}{ ^{n-1}P_3}$

$\sf = \dfrac{^9P_4}{ ^{9-1}P_3}$

$\sf = \dfrac{^9P_4}{ ^{8}P_3}$

$\sf = \dfrac{\dfrac{9!}{(9-4)!}}{\dfrac{(9-1)!}{(9-1-3)!}}$

$\sf = \dfrac{\dfrac{9!}{\cancel{(5)!}}}{\dfrac{8!}{\cancel{(5)!}}}$

$\sf = \dfrac{9!}{8!}$

$\sf = 9$

$\sf RHS = 9$

LHS = RHS

Hence, verified.