Question

1. A committee consisting of 2 mem and 2 women is to be chosen from 5 men and 6 women . In how many ways can this be done ?​

Answer 1

I ve used formula for combinations.

Answer 2

150ways

Step-by-step explanation:

$n! / r!(n-r)!$

2men , 5men

$5 ! / 2 * 1 (3!)$

$5*4*3! / 2*3!$

$20 / 2$

=10

2women , 6women

$6! / 2*1*(4!)$

$6*5*4! / 2*4!$

$30/ 2$

=15

$==> 10*15$

=150 ways