Chapter 9.1, Problem 33E

Solution

To calculate: The numbers of different choices that can be made by the nominating committee if six seniors at Rydell high school meet the qualification for competitive honor scholarship at major university and the university allows the school to nominate up to three candidates, and the school always nominates at least one.

The resultant answer is $41$ .

Given information:

Six seniors at Rydell high school meet the qualification for competitive honor scholarship at major university.

The university allows the school to nominate up to three candidates, and the school always nominates at least one.

Formula used: The number of combinations of n objects taken r at a time denoted by $C_{n_r}$ and given by $C_{n_r}=\frac{n!}{r!\left(n-r\right)!}$ .

Calculation:

The school can choose 1, 2 or 3 students out of 6.

Now the numbers of different choices that can be nominating committee make.

  $C_{6_1}+C_{6_2}+C_{6_3}$

Rewrite the expression using the formula $C_{n_r}=\frac{n!}{r!\left(n-r\right)!}$ ,

The number of ways to selecting 1 senior is:

  $\begin{array}{c}{C}_{6_1}=\frac{6!}{1!\left(6-1\right)!}\\ =\frac{6!}{1!5!}\end{array}$

Now expand the factorial,

  $\frac{6\cdot 5!}{5!}=6$

The number of ways of selecting 2 seniors is:

  $\begin{array}{c}{C}_{6_2}=\frac{6!}{2!\left(6-2\right)!}\\ =\frac{6!}{2!4!}\end{array}$

Now expand the factorial,

  $\frac{6\cdot 5\cdot 4!}{2!4!}=15$

The number of ways of selecting 3 senior is:

  $\begin{array}{c}{C}_{6_3}=\frac{6!}{3!\left(6-3\right)!}\\ =\frac{6!}{3!3!}\end{array}$

Now expand the factorial,

  $\begin{array}{c}\frac{6\cdot 5\cdot 4\cdot 3!}{3!\left(3\cdot 2\cdot 1\right)}=5.4\\ =20\end{array}$

Now, the number of students that can be nominated at most 3,

  $\begin{array}{c}{C}_{6_1}+C_{6_2}+C_{6_3}=6+15+21\\ =41\end{array}$

Therefore, the number of choices which the nominating committee could make is 41.