Question 1055966
So, we have a combination where the order doesn't matter. And we'll pick a group of 3 out of a population of 5:
n = total of 5
k = group of 3
C(5, 3) = n!/((n-k)!k!)
= 5!/((5-3)!3!)
= 5!/(2!*3!)
= 120/12 = 10 possible combinations.