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Question Number 178486 by Acem last updated on 17/Oct/22

$$Reduce\frac{\left(2n\right)!}{1\times 3\times 5\times ...\times (2n-1)}$$

Answered by Rasheed.Sindhi last updated on 17/Oct/22

$$Reduce\frac{\left(2n\right)!}{1\times 3\times 5\times ...\times (2n-1)}$$$$=\frac{\cancel{1}.2.\cancel{3}.4.\cancel{5}...\left(\cancel{2n-1}\right).2n}{\cancel{1}.\cancel{3}.\cancel{5}.....\left(\cancel{2n-1}\right)}$$$$=2.4.6.....2n$$$$=2^n(1.2.3....n)$$$$=2^{n}n!$$

Commented by Acem last updated on 17/Oct/22

$$GoodSir!$$

Commented by Acem last updated on 17/Oct/22

$$Itpreferat2ndlineputttingthefollow$$$$=\frac{2n.\cancel{(2n-1)}(2n-2)\cancel{(2n-3)}(2n-4)....6\times \cancel{5}\times 4\times \cancel{3}\times 2\times \cancel{1}}{\cancel{1\times 3\times 5\times ....\times (2n-1)}}$$$$=2n.(2n-2)(2n-4)..6\times 4\times 2$$$$=2^n.n(n-1)(n-2)...3\times 2\times 1=2^n.n!$$

Commented by Rasheed.Sindhi last updated on 17/Oct/22

$$\mathrm{Ok}\mathrm{sir},\mathrm{thank}\mathrm{you}!$$

Commented by Acem last updated on 17/Oct/22

$$ThebestRasheed,thankyou$$

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