A starting lineup in basketball consists of two guards, two forwards, and a cent

Question

A starting lineup in basketball consists of two guards, two forwards, and a center.

(a) A certain college team has on its roster three centers, four guards, five 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.]
lineups

(b) Now suppose the roster has 3 guards, 5 forwards, 3 centers, and 2 "swing players" (X and Y) who can play either guard or forward. If 5 of the 13 players are randomly selected, what is the probability that they constitute a legitimate starting lineup? (Round your answer to three decimal places.)

Explanation / Answer

If there are x players, y players have to be selected, it can be done in x C y ways, this is called combination.

a) 4G, 5F, 3C and X who can play as F or G.

Above is the superset available.

Need a team of 2F, 2G, 1C. This can be done without X, with X as F, with X as G.

Without X select the 2 forwards, 2 guards and 1 center as follows.

4C2 * 5C2 * 3C1

With X as G, select only 2 forwards, 1 guards and 1 center as follows.

4C1 * 5C2 * 3C1

With X as F, select only 1 forward, 2 guards and 1 center as follows.

4C2 * 5C1 * 3C1

Total number of ways = 390

b)

the team available is 3 G, 5 F, 3 C, x and y who can play F or G. totally 13 players are there.

5 players out of 13 can be selected in 13C5 i.e. 1287 selections

To select a valid team, it can be done with 8 ways

X as F,

X as G,

Y as F,

Y as G,

X and Y as F,

X and Y as G,

X as F and Y as G,

X as G and Y as F

1. X as G can be done by selecting 1 G, 2 F, 1 C among the available (except X and Y). This is the same for Y as G too.

3C1 * 5C2 * 3C1 i.e. 90 for 2 ways

2. X as F can be done by selecting 2 G, 1 F, 1 C among the available (except X and Y). This is the same for Y as F too.

3C2 * 5C1 * 3C1 i.e. 45 for 2 ways

3. X and Y as F can be done by selecting 2 G and 1 C among the available (except X and Y).

3C2 * 3C1 i.e. 9

4. X and Y as G can be done by selecting 2 F and 1 C among the available (except X and Y).

5C2 * 3C1 i.e. 30

5. X as F and Y as G can be done by selecting 1 F, 1 G and 1 C among the available (except X and Y). this is same as X as G and Y as F.

3C1 * 5C1 * 3C1 i.e. 45 for 2 ways

Adding all the 8 ways, we have 180+90+9+30+90 i.e. 399

Probability of selecting a valid team by selecting 5 out of 13 players is 399/1287 i.e. 0.31

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