Question

3(n+1C3)=7(nC2) find n

Answer 1

We are given a question involving Combinatorial Notation. We will be using one simple rule:

$\huge \boxed{\mathsf{\displaystyle ^nC_r = \frac{n!}{(n-r)! \, r!}}}$

Solving, we get the answer as follows:

$\displaystyle 3(^{n+1}C_3)=7.^nC_2 \\ \\ \\\implies 3 \times \frac{(n+1)!}{(n+1-3)! \, 3!}=7\times \frac{n!}{(n-2)! \, 2!} \\ \\\\ \implies 3\times \frac{(n+1)!}{\cancel{(n-2)!} \, 3!}=7\times\frac{n!}{\cancel{(n-2)!} \, 2!} \\ \\ \\\implies \frac{\cancel{3}(n+1)!}{\cancel{6}}=\frac{7n!}{\cancel{2}} \\ \\ \\ \implies \frac{(n+1)!}{n!}=7 \\\\\\\implies n+1=7\\\\\\\implies \huge \boxed{\bold{\mathsf{n=6}}}$

Thus, The value of n is 6.

Answer 2

$\huge\underline\textbf{Answer:-}$

Refer the given attachment:-