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