Question

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

• A. $5$
• B. $8$
• C. $9$
• D. None of these

Answer 1

The correct option is B

$8$

Explanation for the correct option.

Find the solution for the given inequality.

An inequality ${}^{n+1}C_{n-2}-{}^{n+1}C_{n-1}\le 100$ is given.

Solve the given inequality as follows:

$$\begin{gathered} \frac{(n+1)!}{(n-2)!(n+1-n+2)!}-\frac{(n+1)!}{(n-1)!(n+1-n+1)!}\le 100 \\ \Rightarrow \frac{(n+1)n(n-1)\left[(n-2)!\right]}{(n-2)!\left(3\right)!}-\frac{(n+1)n\left[(n-1)!\right]}{(n-1)!\left(2\right)!}\le 100 \\ \Rightarrow \frac{(n+1)n(n-1)}{6}-\frac{(n+1)n}{2}\le 100 \\ \Rightarrow \frac{(n+1)n(n-1)-3(n+1)n}{6}\le 100 \\ \Rightarrow (n+1)n\left[n-1-3\right]\le 600 \\ \Rightarrow (n+1)n\left[n-4\right]\le 600 \end{gathered}$$

Since, $n$ is a positive integer greater than and equal to $2$.

So, the possible values of $n$ are $2,3,4,5,6,7,8$ and $9$.

Therefore, The number of positive integers satisfying the inequality ${}^{n+1}C_{n-2}-{}^{n+1}C_{n-1}\le 100$ is $8$.

Hence, option $B$ is the correct option.