If ${(1 + x)}^{n} = n \sum r = 0 C_{r} x^{r}$ and $\sum \sum_{0 \leq i < j \leq n} C_{i} \times C_{j}$ represent the products of the $C_{i}$ 's taken two at a time, then its value equals

2^{2 n - 1} - \frac{(2 n)!}{{(n!)}^{2}}

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2^{2 n - 1} + \frac{2 n!}{n! n!}

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2^{2 n - 1} - \frac{2 n!}{2 \cdot n! n!}

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None of these

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Solution

The correct option is C $2^{2 n - 1} - \frac{2 n!}{2 \cdot n! n!}$
$A = \sum \sum_{0 \leq i < j \leq 1} C_{i} \times C_{j}$
so, $A = \frac{1}{2} \times ({(\sum_{r = 0}^{n} C_{i})}^{2}) - \sum_{r = 0}^{n} {C_{i}}^{2})$
therefore our required answer is option C.
$\sum_{i = 0}^{n} {{(C_{i})}^{2} =}^{2 n} C_{n}$
${(\sum_{i = 0}^{n} C_{i})}^{2} = 2^{2 n}$

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