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, five 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

a)

No. of different starting lineups = (1 center from 3 center) * (2 gaurd from 4 gaurd) * (2 forward from 5 forward) + (1 center from 3 center) * (2 gaurd from 5 gaurd) * (2 forward from 4 forward)

=3c1*4c2*5c2 + 3c1*5c2*4c2 [for 1st term x is playing as forward and for 2nd term x is playing as gaurd]

=360

2 swings used as guards (g): 2c2 *5c2 *3c1 = 30
2 swings used as forwards (f) : 3c2 *2c2 *3c1 = 9
2 swings used 1g, 1f : 2c1*3c1 *1c1*5c1 *3c1 = 90
1 swing used as g : 2c1*3c1 *5c2 *3c1 = 180
1 swing used as f: 3c2 *2c1*5c1 *3c1 = 90
0 swing used: 3c2 *5c2 *3c1 = 90

legitimate ways = 489
total ways = 13c5 = 1287
Pr = 489/1287 = 0.379

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