A certain college team has on its roster three centers, five guards, three 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.]

Answer: 90 ways

Step-by-step explanation:

Given from full question:The starting line of a basketball has 2 guards, 2 forwards and 1 centre.

Using the expression for combination:

Ck, n = n! / I! (n-k) !

Where n = number of group size

k = number of subset size

N (players sits out) = [3!/1! (3-1) ] * [4! / 2! (3-2) ] * [5! / 2! (5-2) !]

N = [ (3*2) / 1 (2*1) ] * [ (3*2) / (2*1) ] * [ (5*4*3*2) / (2*1 (3*2) ]

N = (6/2) * (6/2) * (120/12)

N=90WAYS