In Exercises 5–36, express all probabilities as fractions.

Soccer Shootout In soccer, a tie at the end of regulation time leads to a shootout by three members from each team. How many ways can 3 players be selected from 11 players available? For 3 selected players, how many ways can they be designated as first, second, and third

Out of 11 players, three are to be selected in any order to play the shootout.

The three players are to be arranged in order of their position.

The concept of combination can be employedtoselect r units from n different units without considering the order of selection.

The following is the formula:

${}^n{C}_r=\frac{n!}{\left(n-r\right)!r!}$

The formula offactorial is used to compute the number of ways n units can be arranged, denoted by !. The formula for factorial is elaborated as follows:

$n!=n\left(n-1\right)\left(n-2\right)...\left(1\right)$

The total number of players is 11.

The number of players to be selected is three.

The selection is made without considering the order of selection.

The number of ways of selecting three players from 11 players is as follows:

$$\begin{gathered} {}^{11}{C}_3=\frac{11!}{\left(11-3\right)!3!} \\ =\frac{11\times 10\times 9}{3\times 2\times 1} \\ =165 \end{gathered}$$

Therefore, the number of ways of selecting three players out of 11 players is equal to 165.

The number of ways three players can be designated as first, second, and third is given as follows:

$$\begin{gathered} 3!=3\times 2\times 1 \\ =6 \end{gathered}$$

Therefore, the number of ways three players can be arranged as first, second, and third equals six.