Question

In a family, 5 males and 3 females are there. In how many ways we can select a group of 2 males and 2 females from the family?

(a) 3

(b) 10

(c) 30

(d) 40

The question was asked by my college director while I was bunking the class.

My query is from Combinations in chapter Permutations and Combinations of Mathematics – Class 11

Answer 1

Right choice is (c) 30

Best explanation: Out of 5 males, 2 males can be selected in ^5C2 ways.

^5C2 = \(\frac{5!}{(5-2)! 2!} = \frac{5*4*3!}{(3)! 2!} = \frac{20}{2}\) = 10 ways.

Out of 3 females, 2 females can be selected in ^3C2 ways .

^3C2 = \(\frac{3!}{(3-2)! 2!} = \frac{3*2!}{(1)! 2!} = \frac{3}{1}\) = 3 ways.

So, by the fundamental principle of counting we can select a group of 2 males and 2 females from the family in 10*3 = 30 ways.