Question

The number of positive integers satisfying the inequality
${}^{n+1}{C}_{n-2}{-}^{n+1}{C}_{n-1}\le 100$ is

• A. One
• B. Eight
• C. Five
• D. None of these

Answer 1

The correct option is B Eight

Given, $\frac{(n+1)!}{3!(n-2)!}-\frac{(n+1)!}{2!.(n-1)!}$
$=\frac{(n-1)\left(n\right)(n+1)}{6}-\frac{n(n+1)}{2}$
$=\left(n\right)(n+1)\frac{n-1-3}{6}$
$\left(n\right)(n+1)\frac{n-4}{6}\le 100$
$\frac{n(n+1)(n-4)}{6}\le 100$
Since $nϵN$
Hence
$2,3,4,5,6,7,8,9$ satisfy the above inequality.