A starting lineup in basketball consists of two guards, two

forwards, and a center.

(a) A certain college team has on its roster 3 centers, 4

guards, 5 forwards, and one individual (X) who can play either

guard or forward. How many different starting lineups can be

created? *Hint:* Consider lineups without X, then lineups

with X as guard, then lineups with X as forward.

(b) Now suppose the roster has 4 guards, 4 forwards, 4 centers, and

2 "swing players" (X and Y) who can play either guard or forward.

If 5 of the 14 players are randomly selected, what is the

probability that they constitute a legitimate starting lineup?

## Answer

## General guidance

Probability: The ratio of the number of favorable outcomes to certain event and total number of possible outcomes is called as the probability of an event.

Let *n *is the total number of observation and *r* is the number of success then the combination is defined as,

The probability of an event is defined as,

## Step-by-step

### Step 1 of 2

**(a)**

The number of different starting lineups is obtained below:

From the given information, the roster has 3 centers, 4 guards, 5 forwards and one individual (*X*) who can play either guard or forward. Moreover, lineup starts with two guards, two forwards and a center.

If *X* is not in lineup, then the possible number ways for lineup is .

If *X* play as forward, then remaining one forward is selected from 5. Therefore, the possible line ups are .

If *X* play as guard, then remaining one guard is selected from 4 guards. Therefore, the possible line ups are .

Thus, the possible different lineups are .

The number of different starting lineups is 390.

The number of different starting lineups is obtained by adding the number of lineups without *X*, the number of lineups with *X* as forward and the number of lineups with *X* as guards.

### Step 2 of 2

**(b)**

Form the given information, 5 players are selected from 14 players randomly. The total number for selecting 5 players is . Moreover, there are 4 guards, 4 forwards, 4 centers and 2 “swing players” (*X* and *Y*) who can play either guard or forward.

The number of lineups without *X* and *Y* is .

If *X* is selected and play as guard, then the possible lineups are .

If *X* is selected and play as forward, then the possible lineups are .

If *Y* is selected and play as guard, then the possible lineups are .

If *Y* is selected and play as forward, then the possible lineups are .

If *X* and *Y* are both selected, here *X* plays as guard and *Y* plays as forward then possible lineups are .

If *X* and *Y* are both selected, here *X* plays as forward and *Y* plays as guard then possible lineups are .

If *X* and *Y* are selected and both plays as guards. Then possible lineups are .

If *X* and *Y* are selected and both plays as forward. Then possible lineups are .

Hence, the possible number of legitimate lineups is,

Therefore, the required probability is,

The probability that the randomly selected players constitute a legitimate starting lineup is 0.3517.

The probability that the randomly selected players constitute a legitimate starting lineup is obtained by dividing the number of possible legitimate lineups by the total number of ways for selecting 5 players.

### Answer

The number of different starting lineups is 390.

The probability that the randomly selected players constitute a legitimate starting lineup is 0.3517.

### Answer only

The number of different starting lineups is 390.

The probability that the randomly selected players constitute a legitimate starting lineup is 0.3517.