In how many different number of ways a combination of 3 men and 2 children can be selected from 6 men and 5 children ?  Choices
A) 20	B) 22	C) 24	D) 26	E) 28
6C3   = 6!/(3! . 3!)   = (6 × 5 × 4)/(3 × 2)   = 5 × 4 = 20.   20)
<result>A</result>
In how many different number of ways a combination of 3 men and 2 children can be selected from 6 men and 5 children ?