How many ways are there to form a committee with 6 members if it must have strictly more women than men?

Question

Suppose that a department contains 8 men and 16 women. How many ways are there to form a committee with 6 members if it must have strictly more women than men?

Answer 1

(1) Chosen 2 men out of 8,  4 women out of 16

      ₈C₂  * ₁₆C₄ = 8!/(6!*2!) * 16!/(12!*4!) 

                 = ((8*7)/2*1 ))* ((16*15*14*13)/(4*3*2*1)

                 = 28 * (2*5*14*13)

                 = 50,960

(2)  Chosen 1 men out of 8, 5 women out of 16

     ₈C₁ * ₁₆C₅ = 8!/(7!*1!) * 16!/(11!*5!) 

                 = 8 * (16*15*14*13*12)/(5*4*3*2*1)

                 = 34,944

(1) + (2),  total 85,904 ways

 

Answer 2

In other word how many ways 2 men can be chosend from a group of 8 and and 4 women from a group of 16

 

⁸P₂ x ¹⁶P₄ = 8 factorial / (8-2)factorial       x      16factorial/(16-4)factorial

 

=> 8x7x6x5x4x3x2x1/6x5x4x3x2x1     x       16x15x14x13x12factorial/12factorial

=>  56 x 524160 = 29352960

 

:)