Question

A starting lineup in basketball consists of two guards, two forwards, and a center.

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?

Answer 1

If there are 14 players in Roster, the number of ways in which 5 of them can be chosen is given by the equation,
14C5 = 2002
Choosing from legitimate players for the positions,
guards: 5C2 = 10
forwards: 4C2 = 6
center: 3C1 = 3
The total number is 10x6x3 = 180
The probability is therefore 180/2002 or equal to 90/1001.