(a) $\lbrack 1.3.5\ldots\ldots(2n - 1)\rbrack 2^{n}$ $= \frac{1.2.3.4.5.6\ldots..(2n - 1)(2n)2^{n}}{2.4.6\ldots..2n}$ $= \frac{(2n)!2^{n}}{2^{n}(1.2.3\ldots.n)} = \frac{(2n)!}{n!}$
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