Chapter 15.7, Problem 20WE

To determine

To find:

The junior and senior class councils each member committees selected from the two class.

Answer to Problem 20WE

  ${}^{10}{C}_3{\times }^{10}{C}_2=5400$

Explanation of Solution

Given:

The junior and senior class councils each member $=10$ .

Concept used:

Using by the formula of the permutation

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

Calculation:

The number of ways of selecting $3$ seniors out of a group of $10$ seniors by the combination

  ${=}^{10}{C}_3$

The number of ways of selecting $2$ seniors out of a group of $10$ seniors by the combination

  ${=}^{10}{C}_2$

Number of committees

  ${=}^{10}{C}_3{\times }^{10}{C}_2$

  $\begin{array}{l}{}^{10}{C}_3{\times }^{10}{C}_2=\frac{10!}{3!\left(10-3\right)!}\times \frac{10!}{2!\left(10-2\right)!}\\ {}^{10}{C}_3{\times }^{10}{C}_2=\frac{10!}{3!7!}\times \frac{10!}{2!8!}\\ {}^{10}{C}_3{\times }^{10}{C}_2=\frac{10\times 9\times 8\times 7!}{3!7!}\times \frac{10\times 9\times 8!}{2!8!}\\ {}^{10}{C}_3{\times }^{10}{C}_2=\frac{10\times 9\times 8}{3\times 2}\times \frac{10\times 9}{2}\\ {}^{10}{C}_3{\times }^{10}{C}_2=10\times 3\times 4\times 5\times 9\\ {}^{10}{C}_3{\times }^{10}{C}_2=120\times 45\\ {}^{10}{C}_3{\times }^{10}{C}_2=5400\end{array}$

Therefore,

The junior and senior class councils each member committees selected from the two classis ${}^{10}{C}_3{\times }^{10}{C}_2=5400$