Question

If ${}^n{P}_3{+}^n{C}_{n-2}=14n,$ then n=

• A. 5
• B. 6
• C. 8
• D. 10

Answer 1

The correct option is A 5

${}^n{P}_3{+}^n{C}_{n-2}=14n$

Expanding

$\frac{n!}{(n-3)!}+\frac{n!}{(n-2)!.2!}=14n$

$2n(n-1)(n-2)+n(n-1)=28n$

$2n^2-6n+4+n-1=28$

$2n^2-5n-25=0$

$(n-5)(2n+5)=0$

-ve not possible n=5