Chapter 9.1, Problem 58E

Solution

To calculate: The numbers of different sets in which 10 players can start when two teams play a game. A basketball team has 13 players and 5 starters without regard to position are chosen by the coach.

The resultant answer is $1,656,369$ ways.

Given information:

A basketball team has 13 players and 5 starters without regard to position are to be chosen by the coach.

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:

There are total 13 basketball players in a team and out of them 5 has to be selected from each team. Then the expression for it will become: $C_{{13}_5}$

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

  $\begin{array}{c}{C}_{{13}_5}=\frac{13!}{5!\left(13-5\right)!}\\ =\frac{13!}{5!8!}\end{array}$

Now expand the factorial,

  $\begin{array}{c}\frac{13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8!}{5!8!}=\frac{13\cdot 12\cdot 11\cdot 10\cdot 9}{5\cdot 4\cdot 3\cdot 2\cdot 1}\\ =1287\end{array}$

Now the number of different set of 10 players that can start when two teams play together is: ${\left(1287\right)}^2=1,656,369$

Therefore, there are $1,656,369$ ways in which 10 players can play.