6. 0/2 points | Previous Answers My Notes o Ask Your Teacher A starting lineup i

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6. 0/2 points | Previous Answers My Notes o Ask Your Teacher 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 as guard, then lineups with X as forward.] 180 X lineups (b) Now suppose the roster has 5 guards, 4 forwards, 3 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? (Round your answer to three decimal places.) Need HelpRead It Talk to a Tutor

Explanation / Answer

(a) The two guards can come in 4C2 ways, the one center in 3C1 ways and the two forwards in 5C2 ways if X is not selected.

Number of teams without X = 4C2 * 3C1 * 5C2 = 6 * 3 * 10 = 180.

If X plays as a guard, the one guard can come in 4C1 ways, the one center in 3C1 ways and the two forwards in 5C2 ways if X is not selected.

Number of teams with X as guard = 4C1 * 3C1 * 5C2 = 4 * 3 * 10 = 120.

If X plays as a forward, the two guards can come in 4C2 ways, the one center in 3C1 ways and the one forward in 5C1 ways if X is not selected.

Number of teams with X as forward = 4C2 * 3C1 * 5C1 = 6 * 3 * 5 = 90.

Total number of teams = 180 + 120 + 90 = 390.

(b) Note that the total number of teams with X as forward or guard was 120 + 90 = 210.

Similarly the number of teams with Y as forward or guard without X = 210.

If X and Y both play, then there are four cases:

(i) X and Y are both guards: The two forwards can come in 5C2 = 10 ways and the one center can come in 3 ways. Total number of teams = 10*3 = 30.

(ii) X and Y are both forwards: The two guards can come in 4C2 = 6 ways and the one center can come in 3 ways. Total number of teams = 6*3 = 18.

(iii) One is a forward and the other is a guard: The one guard can come in 4 ways and the one forward in 5 ways. The one center can come in 3 ways. Total number of teams = 4*5*3 = 60.

Thus the total number of teams with X or Y or both = 210*2 + 30 + 18 + 60*2 = 588.

Total number of ways of choosing 5 players from 14 = 14C5 = 2002.

Probability of a valid selection = 588 / 2002 = 0.294.

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