Question

$C_0{C}_r+C_1{C}_{r+1}+C_2{C}_{r+2}+...+C_{n-r}{C}_n$ is equal to

• A. $\frac{\left(2n\right)!}{\left(n-r\right)!\left(n+r\right)!}$
• B. $\frac{n!}{r!\left(n+r\right)!}$
• C. $\frac{n!}{\left(n+r\right)!}$
• D. None of these

Answer 1

Answer: (A)

$\because {\left(1+x\right)}^n=C_0+C_{1}x+C_2{x}^2+\dots +C_r{x}^r+\dots$
$+C_n{x}^n\dots \left(i\right)$
and ${\left(1+\frac{1}{x}\right)}^n=C_0+C_1\frac{1}{x}+C_2\frac{1}{x^2}+\dots +C_r\frac{1}{x^r}$
$+C_{r+1}\frac{1}{x^{r+1}}+C_{r+2}\frac{1}{x^{r+2}}+\dots +C_n\frac{1}{x^n}\dots \left(ii\right)$
On multiplying $\left(i\right)$ and $\left(ii\right)$, equating coefficient of $x^r$ in $\frac{1}{x^n}{\left(1+x\right)}^{2n}$ or the coefficient of $x^{n+r}$ in ${\left(1+x\right)}^{2n}$, we get the value of required expression which is
${}^{2n}{C}_{n+r}=\frac{\left(2n\right)!}{\left(n-r\right)!\left(n+r\right)!}$