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$C_{0} .^{2 n} C_{n} - C_{1} .^{(2 n - 2)} C_{n} + C_{2} .^{(2 n - 4)} C_{n} \dots =$

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$^{2 n} C_{n}$

$2^{n}$

$2^{2 n}$

$n!$

answer is B.

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Detailed Solution

$C_{0} .^{2 n} C_{n} - C_{1} .^{(2 n - 2)} C_{n} + C_{2} .^{(2 n - 4)} C_{n} \dots$

$consider {({(1 + x)}^{2} - 1)}^{n}$

${({(1 + x)}^{2} - 1)}^{n} =^{n} C_{0} {(1 + x)}^{2 n} -^{n} C_{1} {(1 + x)}^{2 n - 2} +^{n} C_{2} {(1 + x)}^{2 n - 4} + \dots$

$(1 + 2 x + x^{2} - 1) =^{n} C_{0} {(1 + x)}^{2 n} -^{n} C_{1} {(1 + x)}^{2 n - 2} +^{n} C_{2} {(1 + x)}^{2 n - 4} - \dots$

$x^{n} {(2 + x)}^{n} =^{n} C_{0} {(1 + x)}^{2 n} -^{n} C_{1} {(1 + x)}^{2 n - 2} + \dots$

Equating $x^{n}$ coefficient on both sides

$^{n} C_{n} {.2}^{n} =^{n} C_{0} .^{2 n} C_{n} -^{n} C_{1}^{2 n - 2} C_{n} +^{n} C_{2}^{2 n - 4} C_{n} - \dots$

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