Chapter 9.6, Problem 67E

To determine

Find the number of six-member committees which are possible.

Answer to Problem 67E

The number of six-member committees which are possible $292600$

Explanation of Solution

Given:

Since 1 administrator is to be chosen among seven administrators, this can be done in ${}_7{C}_1$

ways.

Similarly, since 3 faculty members are to be chosen from 12 faculty members, this can be done in ${}_{12}{C}_3$ ways.

Since 1 administrator is to be chosen among seven administrators, this can be done in ${}_7{C}_1$

ways.

Similarly, since 3 faculty members are to be chosen from 12 faculty members, this can be done in ${}_{12}{C}_3$ ways.

Since 2 student is to be chosen among 20 students, this can be done in ${}_{20}{C}_2$ ways.

By the fundamental principle of counting, if choosing administrator can be done in

  ${}_7{C}_1$ ways,

faculty members in ${}_{12}{C}_3$ ways, students in ${}_{20}{C}_2$ ways, then total ways of making a 6 -member

team is ${}_7{C}_1{\times }_{12}{C}_3{\times }_{20}{C}_2$

  $\begin{array}{cc}{}_7{C}_1{\times }_{12}{C}_3{\times }_{20}{C}_2=7\times \left(\frac{12!}{(12-3)!3!}\right)\times \left(\frac{20!}{(20-2)!2!}\right)& \\ & =7\times \left(\frac{12\times 11\times 10}{3!}\right)\times \left(\frac{20\times 19}{2!}\right)\\ & =292600\end{array}$