Chapter 11.7, Problem 61S

To determine

The value of the expression $C\left(6,1\right)\cdot C\left(4,1\right)$ .

Answer to Problem 61S

  $24$

Explanation of Solution

Given:

The expression $C\left(6,1\right)\cdot C\left(4,1\right)$ .

Formula used:

  $\begin{array}{l}C\left(n,r\right)=C{}_r=\frac{n!}{r!\cdot \left(n-r\right)!}\\ \text{where,}\\ n!=n\times \left(n-1\right)\times \left(n-2\right)\times \mathrm{...}\times 3\times 2\times 1\end{array}$

Calculation:

In order to evaluate the value of the given expression, use the above combination formula and simplify further as shown below,

  $\begin{array}{c}C\left(6,1\right)\cdot C\left(4,1\right)=C{}_1\cdot C{}_1\\ =\frac{6!}{1!\cdot \left(6-1\right)!}\cdot \frac{4!}{1!\cdot \left(4-1\right)!}\\ =\frac{6!}{1!\cdot 5!}\cdot \frac{4!}{1!\cdot 3!}\\ =\frac{6\cdot 5!}{1\cdot 5!}\cdot \frac{4\cdot 3!}{1\cdot 3!}\\ =\frac{6}{1}\cdot \frac{4}{1}\\ =24\end{array}$

Thus, the value of the expression is $24$.