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Q.
$C_0C_r + C_1C_{r+1} + C_2C_{r+2} + ... + C_{n-r}\, C_n$ is equal to
Binomial Theorem
Solution:
$\because \left(1+x\right)^{n} = C_{0}+C_{1}x+C_{2}x^{2}+\ldots+C_{r}x^{r}+\ldots$
$+C_{n}x^{n}\quad\ldots\left(i\right)$
and $\left(1+\frac{1}{x}\right)^{n} = C_{0}+C_{1} \frac{1}{x}+C_{2} \frac{1}{x^{2}}+\ldots+C_{r} \frac{1}{x^{r}}$
$+ C_{r+1} \frac{1}{x^{r+1}}+ C_{r+2} \frac{1}{x^{r+2}}+ \ldots+ C_{n} \frac{1}{x^{n}}\quad\ldots\left(ii\right)$
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
$^{2n}C_{n+r} = \frac{\left(2n\right)!}{\left(n-r\right)!\left(n+r\right)!}$