Hello, mathh18!

With no constraints, there are: . possible committees.There are 9 men, 12 women in the math club.

How many possible committees of four people can be formed

if Jill (woman) and Jack (man) cannot be on the same committee?

How many committtees haveJack and Jill?both

Since Jack and Jill are on the committee, choose 2 more from the other 19 people.

. . There are: . committees with both Jack and Jill.

Therefore, there are: . committees

. . in which Jack and Jill do not serve together.

There are two cases to consider:How many four-person committees have more women than men

if Jill and Jack cannot be on the same committee?

. . (1) 4 women

. . (2) 3 women, 1 man

(1) 4 women

. . There are: . all-women committees.

(2) 3 women, 1 man

There are two cases to consider:

. . (a) Jill isthe committee.on

. . (b) Jill ison the committee.not

(a) Jill isthe committee.on

Choose 2 more women from the other 11 women: . ways.

Choose one man from the 8 men (other than Jack): . ways.

Hence, there are: . committees with 3 women, 1 man (with Jill).

(b) Jill ison the committee.not

Choose 3 women from the other 11 women: . ways.

Choose 1 man from the available 9 men: . ways.

Hence, there are: . committees with 3 women, 1 man (without Jill).

Therefore, there are: . committees

. . which have more women than men, and Jack and Jill do not serve together.