Question

Value of
$S=\frac{1}{m!}{C}_0+\frac{n}{(m+1)!}{C}_1+\frac{n(n-1)}{(m+2)!}{C}_2+.....+\frac{n(n-1)...2.1}{(m+n)!}$ is

• A. $\frac{(m+n+1)(m+n+2)....(m+2n)}{(m+n)!}$
• B. ${{}^{m+n}C}_n$
• C. $\frac{1}{(m+n)!}\left({{}^{m+n}C}_n\right)$
• D. $0$

Answer 1

Answer: (A)

$\frac{(m+n+1)(m+n+2)....(m+2n)}{(m+n)!}$

Simplifying the above expression we get
$S=\frac{n!}{(m+n)!}[{}^{m+n}{C}_m.{}^n{C}_0+{}^{m+n}{C}_{m+1}.{}^n{C}_1+...+{}^{m+n}{C}_{m+n}.{}^n{C}_n]$
Now ${}^n{C}_r={}^n{C}_{n-r}$
Hence $S=\frac{n!}{(m+n)!}[{}^{m+n}{C}_m.{}^n{C}_n+{}^{m+n}{C}_{m+1}.{}^n{C}_{n-1}+...+{}^{m+n}{C}_{m+n}.{}^n{C}_0]$
$=\frac{n!}{(m+n)!}.{}^{m+2n}{C}_{m+n}$
$=\frac{n!}{(m+n)!}.\frac{(m+2n)!}{n!.(m+n)!}$
$=\frac{(m+2n)!}{(m+n){!}^2}$