Q. 1.7 - A First Course in Probability

Question

Give a combinatorial explanation of the identity

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

Step-by-Step Solution

Answer

The given identity $\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ n-r\end{array}\right)$ can be explained as the no. of possible combinations of selecting r objects from n objects is equivalent to selecting $n-r$ objects from $n$ objects.

Step 1. Given information.

The given identity is 

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

and we have to give its combinatorial explanation.

We know that $\left(\begin{array}{c}x\\ r\end{array}\right)=\frac{x!}{r!\left(x-r\right)!}$.

Step 2. Give combinatorial explanation of the given identity.

Taking L.H.S of the given identity, we have 

$\left(\begin{array}{c}n\\ r\end{array}\right)$, which can be written as  $\frac{n!}{r!\left(n-r\right)!}$.

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

Taking R.H.S of the given identity, we have $\left(\begin{array}{c}n\\ n-r\end{array}\right)$, which can be written as $\frac{n!}{\left(n-r\right)!(n-n+r)!}=\frac{n!}{r!\left(n-r\right)!}$

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

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

The given identity $\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ n-r\end{array}\right)$ can be explained as the no. of possible combinations of selecting $r$ objects from $n$ objects is equivalent to selecting $n-r$ objects from $n$ objects.