The value of $C_{1}^{2} + C_{2}^{2} . . . . + C_{n}^{2}$ (where $C_{i}$ is the $i^{t h}$ coefficient of $(1 + x)^{n}$ expansion), is:

\frac{n^{n}}{n!}

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\frac{2 n!}{n! n!}

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\frac{2 n!}{n!}

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\frac{n! \times 2^{n}}{2 n!}

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Solution

The correct option is B $\frac{2 n!}{n! n!}$
${(1 + x)}^{n} =^{n} C_{0} + x (^{n} C_{1}) + x^{2} (^{n} C_{2}) + x^{3} (^{n} C_{3}) + . . . + x^{n} (^{n} C_{n})$
${(x + 1)}^{n} = x^{n} (^{n} C_{0}) + x^{n - 1} (^{n} C_{1}) + . . . + x^{0} (^{n} C_{n})$
Now, $\sum_{r = 0}^{n} {(^{n} C_{r})}^{2}$ = co-efficient of $x^{n}$ in $(1 + x)^{2 n} =^{2 n} C_{n}$
Thus, $^{2 n} C_{n} = {(^{n} C_{0})}^{2} + {(^{n} C_{1})}^{2} + . . . {(^{n} C_{n})}^{2}$ , where $(^{n} C_{i})$ is the $i^{t h}$ coefficient of $(1 + x)^{n}$ expansion.
So, $C_{1}^{2} + C_{2}^{2} + . . . + C_{n}^{2} = (^{2 n} C_{n}) = \frac{2 n!}{n! n!}$

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