Chapter 9.1, Problem 45E

Solution

To identify: The number of different lunches possible if a lunch consists of an entrée, two vegetables, and a dessert. There are four entrees, six vegetables, and six desserts.

The options are:

  $\left(a\right)16$

  $\left(b\right)25$

  $\left(c\right)144$

  $\left(d\right)360$

  $\left(e\right)720$

The correct option is $\left(d\right)360$ .

Given information:

There are four entrees, six vegetables, and six desserts.

Explanation:

There are four entrees, six vegetables, and six desserts from which an entrée, two vegetables, and a dessert are to be selected to make a lunch.

Then the expression for it will become: $C_{4_1}\times C_{6_2}\times C_{6_1}$

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

  $C_{4_1}\times C_{6_2}\times C_{6_1}=\frac{4!}{1!\left(4-1\right)!}\times \frac{6!}{2!\left(6-2\right)!}\times \frac{6!}{1!\left(6-1\right)!}$

Now expand the factorials,

  $\begin{array}{c}\frac{4!}{1!3!}\times \frac{6!}{2!4!}\times \frac{6!}{1!5!}=\frac{4\cdot 3!}{1!3!}\times \frac{6\cdot 5\cdot 4!}{2!4!}\times \frac{6\cdot 5!}{1!5!}\\ =4\times 15\times 6\\ =360\end{array}$

Therefore, there are 360 ways in which the lunch can be made. Option $\left(d\right)360$ should be the correct answer.