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, four 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 5 guards, 5 forwards, 3 centers, and 2 “swing players” (X and Y) who can play either guard or forward. If 5 of the 15 players are randomly selected, what is the probability that they constitute a legitimate starting lineup?
Answer Step 1 of 2 a) There are three possibilities: x is not picked, x is picked as a forward, and x is picked as a guard. When x is not picked Number of different lineups= (4 2)(4 2)(3 1)=108 c c c when x is picked as a forward, you only need to pick one more forward, Number of different lineups= (4 1)(4 2)c 1)c2 c the same number will result when you pick x as a guard: 72 By adding all three results number of lineups=108+72+72=252