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.] 180 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 Help? LReadIt-1 Talk to a Tutor!!

Explanation / Answer

Basketball has 2G, 2F and 1C

a. roster has 3C. 4G. 5F. and X who can plat either G. or F.

Lineup if X plays G - 5C2*3C1*5C2 = 10*3*10 = 300

Line up if X plays F - 6C2*5C2*3C1 = 15*10*3 = 450

So, total lineups possible is 300+450 = 750

b. total combos = 14C5

Total legit combos = ?

XY both play G - 7C2*4C2*3C1 = 378
XY both play F - 5C2*6C2*3C1 = 450
X plays G, Y plays F and vice-versa - 2*6C2*5C2*3C1 ( so its counted twice) = 900

So, total legit combos = 378+450+900 = 1728

Hence, probabity that they consitute a legit starting line up = 1728/14C5 = .8631

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