Question

Give an analytic proof of Equation (4.1).

Short answer

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

Step-by-step solution

Step 1. Given information.

We have to prove the combinatorial explanation of the identity, that is$\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ n-r\end{array}\right)$.

Step 2. Prove the combinatorial explanation of the identity.

The combinatorial explanation of the identity is $\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ n-r\end{array}\right)$.

Taking L.H.S, we have $\left(\begin{array}{c}n\\ r\end{array}\right)$.

Since role="math" localid="1647855220565" $\left(\begin{array}{c}x\\ r\end{array}\right)=\frac{x!}{r!(x-r)!}$, so

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

Therefore, L.H.S = role="math" localid="1647855382907" $\frac{n!}{r!\left(n-r\right)!}$

Taking R.H.S, we have $\left(\begin{array}{c}n\\ n-r\end{array}\right)$

Since $\left(\begin{array}{c}x\\ r\end{array}\right)=\frac{x!}{r!(x-r)!}$, so

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

Therefore, R.H.S $=\frac{n!}{r!\left(n-r\right)!}$

As L.H.S = R.H.S, hence it is proved that$\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ n-r\end{array}\right)$.