If (2n)!=x.2n{1.3.5…..(2n−1)} then x=

2n!=2n2n-12n-22n-3····3·2·1 =2nnn-1n-2···3·2·1·2n-12n-3···3·1 =2nn!2n-12n-32n-5··3·1Therefore, x=n!

If (2n)!=x.2n{1.3.5…..(2n−1)} then x=

2n!=2n2n-12n-22n-3····3·2·1 =2nnn-1n-2···3·2·1·2n-12n-3···3·1 =2nn!2n-12n-32n-5··3·1Therefore, x=n!

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If $(2 n)! = x {.2}^{n} {1.3.5 \dots . . (2 n - 1)}$ then x=

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$(2 n)!$

$2 (n - 1)!$

$n!$

$(n - 1)!$

answer is A.

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Detailed Solution

$\begin{array}{rcl} (2 n)! & = & (2 n) (2 n - 1) (2 n - 2) (2 n - 3) \cdot \cdot \cdot \cdot 3 \cdot 2 \cdot 1 \\ = & 2^{n} (n) (n - 1) (n - 2) \cdot \cdot \cdot 3 \cdot 2 \cdot 1 \cdot (2 n - 1) (2 n - 3) \cdot \cdot \cdot 3 \cdot 1 \\ = & 2^{n} (n!) [(2 n - 1) (2 n - 3) (2 n - 5) \cdot \cdot 3 \cdot 1] \end{array}$