Question

If ${}^{n+1}{C}_4=9^n{C}_2$, then $n=$

• A. $10$
• B. $9$
• C. $12$
• D. $11$

Answer 1

The correct option is D $11$

${}^{n+1}{C}_4=9{}^n{C}_2$

$\frac{(n+1)!}{(n-3)14!}=9\frac{n!}{(n-2)!2!}$

$\frac{(n+1)n!}{(n-3)!4!}=\frac{9n!}{(n-2)(n-3)!2!}$

$\frac{n+1}{4\times 3\times 2!}=\frac{9}{(n-2)2!}$

$(n+1)(n-2)=12\times 9$

$n^2-2n+n-2=108$

$n^2-n-2=108$

$n^2-n=110$

$n^2-n-110=0$

On simiplying, we get

$n=11$

$\because n^2-11n+10n-110=0$

$n(n-11)+10(n-11)=0$

$(n+10)(n-11)=0$

$n=11,-10$

$\because n\ne -10$

$\Rightarrow n=11$