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.] 390 lineups (b) Now suppose the roster has 3 guards, 4 forwards 3 centers, and 2 "swing players" (X and Y) who can play either guard or forward. If 5 of the 12 players are randomly selected, what is the probability that they constitute a legitimate starting lineup? (Round your answer to three decimal places.)

Explanation / Answer

Solution

Back-up Theory

Out n things, r things can be selected in nCr = (n!)/{(r!)(n - r)!}.

Now, to work out the solution,

Composition of legitimate line-up: 2 Guards + 2 Forwards + 1 Center

Part (a)

Availablilty

Case 1: 5 Guards(X included) + 5 Forwards + 3 Centers

2 Guards out of 5 + 2 Forwards out of 5 + 1 Center out of 3 can selected in (5C2)x(5C2)x(3C1) ways

= 10x10x3 = 300

Case 2: 4 Guards + 6 Forwards(X included) + 3 Centers

2 Guards out of 4 + 2 Forwards out of 6 + 1 Center out of 3 can selected in (4C2)x(6C2)x(3C1) ways

= 6x15x3 = 270

Case 3: X is not included : 4 Guards + 5 Forwards + 3 Centers

2 Guards out of 4 + 2 Forwards out of 5 + 1 Center out of 3 can selected in (4C2)x(5C2)x(3C1) ways

= 6x10x3 = 180

Adding case 1, case 2 and case 3 possibilities, total number of legitimate starting line-up possibilities

= 750 ANSWER

Part (b)

Availablilty

Case 1: 5 Guards(X and Y included) + 4 Forwards + 3 Centers

2 Guards out of 5 + 2 Forwards out of 4 + 1 Center out of 3 can selected in (5C2)x(4C2)x(3C1) ways

= 10x6x3 = 180

Case 2: 4 Guards(X included) + 5 Forwards(Y included) + 3 Centers

2 Guards out of 4 + 2 Forwards out of 5 + 1 Center out of 3 can selected in (4C2)x(5C2)x(3C1) ways

= 6x10x3 = 180

Case 3: 4 Guards(Y included) + 5 Forwards(X included) + 3 Centers

2 Guards out of 4 + 2 Forwards out of 5 + 1 Center out of 3 can selected in (4C2)x(5C2)x(3C1) ways

= 6x10x3 = 180

Case 4: 3 Guards + 6 (X and Y included)Forwards + 3 Centers

2 Guards out of 3 + 2 Forwards out of 6 + 1 Center out of 3 can selected in (3C2)x(6C2)x(3C1) ways

= 3x15x3 = 135

Case 5: Neither X nor Y is included

3 Guards + 4 (X and Y included)Forwards + 3 Centers

2 Guards out of 3 + 2 Forwards out of 4 + 1 Center out of 3 can selected in (3C2)x(4C2)x(3C1) ways

= 3x6x3 = 54

Thus, there are {(3 x 180) + 135 + 54} = 729 legitimate starting line-up possibilities.

Out of 3 Guards + 4 Forwards + 3 Centers + 2 swing palyers (X and Y), a team of 5, without any restriction, can be formed in 12C5 = 792 ways.

So, probability a randomly picked team will be legitimate line-up = 729/792 = 0.9205 ANSWER

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