$\begin{array}{1 1} 6 \\ 7 \\ 8 \\ 9 \end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

$^{n+1}C_{n-2}-^{n+1}C_{n-1}\leq 50$

$\Rightarrow\:\large\frac{(n+1)!}{(n-2)!.3!}-\frac{(n+1)!}{(n-1)!.2!}$$\leq 50$

$\Rightarrow\:\large\frac{(n+1)n}{2}$$\bigg[\large\frac{n-1}{3}$$-1\bigg]\leq 50$

$\Rightarrow\:(n+1)n(n-4)\leq 300$

Since $^{n+1}C_{n-2}$ exists for $n\geq 2$,

and for $n=9$, $(n+1)n(n-4)=450\nleq 300$, $n<9$

$i.e., n=2,3,4,5,6,7,8$

$\therefore$ The required no. of positive integers = 7

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...