Chapter 6.1, Problem 44E

Solution

To verify: the given identity.

  $C{}_r\cdot C{}_m=C{}_m\cdot C{}_{r-m}$.

Given:

  $C{}_r\cdot C{}_m=C{}_m\cdot C{}_{r-m}$.

Formula used:

  $C{}_r=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

Consider,

  $\begin{array}{c}C{}_r\cdot C{}_m=\frac{n!}{r!\left(n-r\right)!}\cdot \frac{r!}{m!\left(r-m\right)!}\\ C{}_r\cdot C{}_m=\frac{n!}{\cancel{r!}\left(n-r\right)!}\cdot \frac{\cancel{r!}}{m!\left(r-m\right)!}\\ C{}_r\cdot C{}_m=\frac{n!}{m!\left(n-r\right)!\left(r-m\right)!}\to \left(1\right)\end{array}$

And,

  $\begin{array}{c}C{}_m\cdot C{}_{r-m}=\frac{n!}{m!\left(n-m\right)!}\cdot \frac{\left(n-m\right)!}{\left(r-m\right)!\left[\left(n-m\right)-\left(r-m\right)\right]!}\\ C{}_m\cdot C{}_{r-m}=\frac{n!}{m!\cancel{\left(n-m\right)!}}\cdot \frac{\cancel{\left(n-m\right)!}}{\left(r-m\right)!\left(n-r\right)!}\\ C{}_m\cdot C{}_{r-m}=\frac{n!}{m!\left(n-r\right)!\left(r-m\right)!}\to \left(2\right)\end{array}$

From the equations (1) and (2), it follows that

  $C{}_r\cdot C{}_m=C{}_m\cdot C{}_{r-m}$