Question

Present a combinatorial explanation of why

$\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ r,n-r\end{array}\right)$

Short answer

It is proved that$\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ r,n-r\end{array}\right)$

Step-by-step solution

Step 1. Given information.

It is given that$\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ r,n-r\end{array}\right)$.

Step 2. Give explanation for nr=nr, n-r

$\left(\begin{array}{c}n\\ r\end{array}\right)$represents the no. of ways in which $r$objects can be selected from $n$objects.

When this is done, $r$objects that are selected are grouped separately and $\left(n-r\right)$objects that are not selected are grouped separately.

Hence, this is equivalent to separating $n$objects into two groups, one having $r$objects and another having $\left(n-r\right)$objects.

This can be done in$\frac{n!}{r!(n-r)!}\text{ways}=\left(\begin{array}{c}n\\ r,n-r\end{array}\right)$

Therefore,$\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ r,n-r\end{array}\right)$.