Hey, thanks in advance.
5 people are to be chosen from 4 male and 6 females, how many possible committees can be selected which include more females than males?
I will give this a go!.
1 - First scenario: You want a committee of 5 people 4 of which must be female and 1 male:
6C4 x 4C1 =15 x 4 = 60 ways of choosing in this scenario.
2- Second scenario: You want a committee of 5 people 3 of which must be female and 2 male:
6C3 x 4C2 =20 x 6 =120 ways of choosing in this scenario.
So, the total of the two scenarios =60 + 120 = 180 ways of choosing a committee of 5 people.
Note: Somebody should check this.
