Chapter 9.2, Problem 31E

Solution

To prove: The $\left(\begin{array}{l}n\\ r\end{array}\right)=\left(\begin{array}{l}n-1\\ r-1\end{array}\right)+\left(\begin{array}{l}n-r\\ r\end{array}\right)$ by using formula $\left(\begin{array}{l}n\\ r\end{array}\right)=\frac{n!}{r!\left(n-r\right)!}$ for all integers.

Given information:

The given binomial

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

Proof:

Prove the statement as follows:

  $\begin{array}{c}\left(\begin{array}{l}n-1\\ r-1\end{array}\right)+\left(\begin{array}{l}n-1\\ r\end{array}\right)=\frac{\left(n-1\right)!}{r!\left(\left(n-1\right)-r\right)!}+\frac{\left(n-1\right)!}{\left(r-1\right)!\left(\left(n-1\right)-\left(r-1\right)\right)!}\\ =\frac{\left(n-1\right)!}{r!\left(n-r-1\right)!}+\frac{\left(n-1\right)!}{\left(r-1\right)!\left(n-r\right)!}\\ =\frac{\left(n-r\right)\left(n-1\right)!}{r!\left(n-r\right)\left(n-r-1\right)!}+\frac{r\left(n-1\right)!}{r\left(r-1\right)!\left(n-r\right)!}\left(\begin{array}{c}n!=n\cdot \left(n-1\right)\cdot \left(n-2\right)\cdot \mathrm{......}\cdot 1\\ =n\cdot \left(n-1\right)!\end{array}\right)\\ =\frac{\left(n-r\right)\left(n-1\right)!}{r!\left(n-r\right)!}+\frac{r\left(n-1\right)!}{r!\left(n-r\right)!}\\ =\frac{\left(n-1\right)!}{r!\left(n-r\right)!}\left(\left(n-r\right)+r\right)\\ =\frac{\left(n-1\right)!}{r!\left(n-r\right)!}\left(n\right)\\ =\frac{n\left(n-1\right)!}{r!\left(n-r\right)!}\\ =\frac{n!}{\begin{array}{l}r!\left(n-r\right)!\\ =\left(\begin{array}{l}n\\ r\end{array}\right)\end{array}}\end{array}$.

So, proof that all integers is

  $\left(\begin{array}{l}n\\ r\end{array}\right)=\left(\begin{array}{l}n-1\\ r-1\end{array}\right)+\left(\begin{array}{l}n-1\\ r\end{array}\right)$ .