Chapter 9.1, Problem 48E

Solution

To identify: The number of different ways in which members can fill out the ballot if 3 new board members are to be elected by voting. There are 5 candidates and the members can check off all the names of the candidates or none of them.

The options are:

  $\left(a\right)10$

  $\left(b\right)20$

  $\left(c\right)32$

  $\left(d\right)125$

  $\left(e\right)243$

The correct option is $\left(c\right)32$ .

Given information:

There are 5 candidates from which 3 candidates with most checks are to be selected and the members can check off all the names of the candidates or none of them.

Explanation:

Total number of candidates: 5

The number of ways for filling the ballot for none of the candidates: $C_{5_0}$

The number of ways filling the ballot for only one of the candidates: $C_{5_1}$

The number of ways filling the ballot for two candidates: $C_{5_2}$

The number of ways filling the ballot for three candidates: $C_{5_3}$

The number of ways filling the ballot for four candidates: $C_{5_4}$

The number of ways filling the ballot for five candidates: $C_{5_5}$

The total number of ways will become: $C_{5_0}+C_{5_1}+C_{5_2}+C_{5_3}+C_{5_4}+C_{5_5}$

Now solve it:

  $\begin{array}{l}=\frac{5!}{0!\left(5-0\right)!}+\frac{5!}{1!\left(5-1\right)!}+\frac{5!}{2!\left(5-2\right)!}+\frac{5!}{3!\left(5-3\right)!}+\frac{5!}{4!\left(5-4\right)!}+\frac{5!}{5!\left(5-5\right)!}\\ =\frac{5!}{0!5!}+\frac{5!}{1!4!}+\frac{5!}{2!3!}+\frac{5!}{3!2!}+\frac{5!}{4!1!}+\frac{5!}{5!0!}\\ =1+5+10+10+5+1\\ =32\end{array}$

Then the total number of ways for filling the ballot will become: 32

Therefore, Option $\left(c\right)32$ should be the correct answer.