Mathematics - Binomial Theorem Question with Solution | TestHub

For $r\ge 0,\frac{1}{r+1}{C}_r^2=\frac{1}{r+1}\frac{n!}{r!(n-r)!}{C}_r=\frac{1}{n+1}\frac{(n+1)!}{(r+1)!(n-r)!}{C}_r$

$$=\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{n+1}{r+1}\right)\left(\genfrac{}{}{0ex}{}{n}{r}\right)$$

Thus, $S=\frac{1}{n+1}{S}_1$ where $S_1={\sum }_{r=0}^n\left(\genfrac{}{}{0ex}{}{n+1}{n-r}\right)\left(\genfrac{}{}{0ex}{}{n}{r}\right)$

= number of ways of selecting $n$ persons out of ( $n+1$ ) men and $n$ women

= number of ways of selecting $n$ persons out of $(2n+1)$ persons

$=\left(\genfrac{}{}{0ex}{}{2n+1}{n}\right)=\frac{(2n+1)!}{n!(n+1)!}$

$$\therefore S=\frac{1}{n+1}\frac{(2n+1)!}{n!(n+1)!}=\frac{(2n+1)!}{(n+1){!}^2}=\frac{1}{(2n+2)}\left(\genfrac{}{}{0ex}{}{2n+2}{n+1}\right)$$