$C_0 C_r+C_1 C_{r+1}+C_2 C_{r+2}+\ldots+C_{n-r} C_n$ is equal to
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$C_0 C_r+C_1 C_{r+1}+C_2 C_{r+2}+\ldots+C_{n-r} C_n$ is equal to

(A) $\frac{(2 n) !}{(n-r) !(n+r) !}$

(B) $\frac{n !}{(-r) !(n+r) !}$

(C) $\frac{n !}{(n-r) !}$

(D) None of these

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Best answer

SOLUTION —

$\because(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_r x^r+\ldots$

and $\left(1+\frac{1}{x}\right)^n=C_0+C_1 \frac{1}{x}+C_2 \frac{1}{x^2}+\ldots C_{r+1} \frac{1}{x^{r+1}}$$+C_{r+2} \frac{1}{x^{r+2}} \ldots C_n \frac{1}{x^n}$

Multiplying Eqs. (i) and (ii) and equating coefficient of $x^r$ in $\frac{1}{x^n}(1+x)^{2 n}$ or the coefficient of $x^{n+r}$ in $(1+x)^{2 n}$, we get the value of required expression

$={ }^{2 n} C_{n+r}=\frac{(2 n) !}{(n-r) !(n+r) !}$

So, The correct option will be (A).

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