# To explain: why the combination binom{n}{r}=binom{n}{r,n-r}.

To explain: why the combination $\left(\genfrac{}{}{0}{}{n}{r}\right)=\left(\genfrac{}{}{0}{}{n}{r,n-r}\right)$

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Formula used:
The number of combinations of choosing r objects from n objects is
${}^{\left\{n\right\}}{C}_{r}=\frac{n!}{\left(n-r\right)r!}$
The number of combinations of choosing two groups x and y from n objects is
${}^{\left\{n\right\}}{C}_{x,y}=\frac{n!}{x!y!\left(n-x-y\right)!}$
Consider the number of combinations of choosing r objects from n objects $=\left(\genfrac{}{}{0}{}{n}{r}\right)$
Let n is divided in to two groups one is having r objects and other group is having n
$\left(n-r\right)$ Objects, that is $n=r+\left(n-r\right)$, then
The number of ways of Choosing two groups r and from n objects
$\left(\genfrac{}{}{0}{}{n}{r,n-r}\right)=\frac{n!}{r!\left(n-r\right)!\left(n-\left(n-r\right)-r\right)!}$
$=\frac{n!}{r!\left(n-r\right)\left(n-n+r-r\right)}$
$=\frac{n!}{r!\left(n-r\right)!}$
$=\left(\genfrac{}{}{0}{}{n}{r}\right)$
Therefore, $\left(\genfrac{}{}{0}{}{n}{r}\right)=\left(\genfrac{}{}{0}{}{n}{r,n-r}\right)$
Hence, explained