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
