Question

If ${}^n{p}_4=20{\times }^n{p}_2$ then find the value n.

• A. $7$
• B. $5$
• C. $6$
• D. $4$

Answer 1

The correct option is A $7$

Given that ${}^n{P}_4{=}^n{P}_2\times 20$,

$⟹\frac{n!}{(n-4)!}=20\times \frac{n!}{(n-2)!}⟹1=\frac{20}{(n-2)(n-3)}⟹(n-2)(n-3)=20⟹n^2-5n-14=0⟹n=7$,

As $n>0$