Question

If ${}^{n+1}{C}_3={2.}^n{C}_2,$ then =

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

The correct option is C

5

${}^{n+1}{C}_3={2.}^n{C}_2$

$\Rightarrow \frac{(n+1)!}{3!(n-2)!}=2\times \frac{n!}{2!(n-2)!}$

$\Rightarrow \frac{(n+1)n!}{3\times 2!(n-2)!}=2\times \frac{n!}{2!(n-2)!}$

$\Rightarrow n+1=6$

$\Rightarrow n=5$