Chapter 8.5, Problem 14E

To determine

To find: the numbers of schedules are possible.

Answer to Problem 14E

Number of schedules is 30.

Explanation of Solution

Given information:

A college student is preparing a course schedule for the next semester. The student select one of two mathematics course, one of three science course, and one of five courses from the social sciences and humanities.

Calculation:

Since 1 mathematics course is to be chosen among 2 mathematics courses, this can be done in ${}_2{C}_1$ ways.

Similarly, since 1 science course is to be chosen from 3 science courses, this can done in ${}_3{C}_1$ ways.

Since 1 social science and humanities course is to be chosen among 5 social science and humanities courses, this can be done in ${}_5{C}_1$ ways.

Fundamental principle of counting

Let $E_1$ and $E_2$ be two events. The first event $E_1$ can occur in $m_1$ different ways. After $E_1$ has occurred, $E_2$ can occur in $m_2$ different ways. The number of ways that the two events occur is $m_1\cdot m_2$

By the fundamental principle of counting, if choosing mathematics course can be done in ${}_2{C}_1$ ways, science course in ${}_3{C}_1$ ways, science and humanities course in ${}_5{C}_1$ ways, then total ways of making a 6 member team is ${}_2{C}_1{\times }_3{C}_1{\times }_5{C}_1$ .

  $\begin{array}{c}{}_2{C}_1{\times }_3{C}_1{\times }_5{C}_1=\left(\frac{2!}{\left(2-1\right)!1!}\right)\times \left(\frac{3!}{\left(3-1\right)!1!}\right)\times \left(\frac{5!}{\left(5-1\right)!1!}\right)\\ =\left(\frac{2}{1}\right)\times \left(\frac{3!}{2!}\right)\times \left(\frac{5}{1!}\right)\\ =30\end{array}$