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Question-4075




Question Number 4075 by Yozzii last updated on 27/Dec/15
Answered by prakash jain last updated on 28/Dec/15
^(2r) C_r =((2r!)/(r!r!))  2^r ∙r!=2∙4∙6∙...∙2r  2r!=1∙2∙3∙4∙...∙(2r−1)∙2r  ^(2r) C_r =((1∙3∙5∙7∙...(2r−1))/(r!))2^r
$$\:^{\mathrm{2}{r}} {C}_{{r}} =\frac{\mathrm{2}{r}!}{{r}!{r}!} \\ $$$$\mathrm{2}^{{r}} \centerdot{r}!=\mathrm{2}\centerdot\mathrm{4}\centerdot\mathrm{6}\centerdot…\centerdot\mathrm{2}{r} \\ $$$$\mathrm{2}{r}!=\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}\centerdot…\centerdot\left(\mathrm{2}{r}−\mathrm{1}\right)\centerdot\mathrm{2}{r} \\ $$$$\:^{\mathrm{2}{r}} {C}_{{r}} =\frac{\mathrm{1}\centerdot\mathrm{3}\centerdot\mathrm{5}\centerdot\mathrm{7}\centerdot…\left(\mathrm{2}{r}−\mathrm{1}\right)}{{r}!}\mathrm{2}^{{r}} \\ $$

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