Chapter 6.1, Problem 5GP

Solution

To calculate: The number of sets for exactly 3 tragedies and 2 histories.

5040

Given information:

Total number of play is 38 in which 18 is comedy, 10 is historical and 10 is tragedy,

Formula used:

Formula of selecting $r$ objects from $n$ distinct objects is $\text{C}{}_{\text{r}}=\frac{n!}{\left(n-r\right)!r!}$ .

Calculation:

Choose 3 from 10 tragedies.

  $\begin{array}{c}\text{C}{}_3=\frac{10!}{\left(10-3\right)!3!}\\ =\frac{10!}{7!3!}\end{array}$

Choose 2 from 10 histories.

  $\begin{array}{c}\text{C}{}_2=\frac{10!}{\left(10-2\right)!3!}\\ =\frac{10!}{8!2!}\end{array}$

Now, calculate number of different possible sets

  $\begin{array}{c}\text{C}{}_3\cdot \text{C}{}_2=\frac{10!}{7!3!}\cdot \frac{10!}{8!2!}\\ =\frac{720}{6}\cdot \frac{90}{2}\\ =5400\end{array}$

Hence, total number of such sets is 5040.