# probability/combinations & binomial theory

• Apr 4th 2010, 07:01 PM
mathh18
probability/combinations & binomial theory
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 four person committees have more women than men if Jill and Jack cannot be on the same committee?
• Apr 4th 2010, 08:31 PM
Soroban
Hello, mathh18!

Quote:

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?

With no constraints, there are: .$\displaystyle {21\choose4} \,=\,5985$ possible committees.

How many committtees have both Jack and Jill?
Since Jack and Jill are on the committee, choose 2 more from the other 19 people.
. . There are: .$\displaystyle {19\choose2} \,=\,171$ committees with both Jack and Jill.

Therefore, there are: .$\displaystyle 5985 - 171 \:=\:\boxed{5814}$ committees
. . in which Jack and Jill do not serve together.

Quote:

How many four-person committees have more women than men
if Jill and Jack cannot be on the same committee?

There are two cases to consider:
. . (1) 4 women
. . (2) 3 women, 1 man

(1) 4 women
. . There are: .$\displaystyle {12\choose4} \:=\:{\color{blue}495}$ all-women committees.

(2) 3 women, 1 man

There are two cases to consider:
. . (a) Jill is on the committee.
. . (b) Jill is not on the committee.

(a) Jill is on the committee.
Choose 2 more women from the other 11 women: .$\displaystyle {11\choose2} \,=\,55$ ways.
Choose one man from the 8 men (other than Jack): .$\displaystyle {8\choose1} \,=\,8$ ways.

Hence, there are: .$\displaystyle 55\cdot8 \:=\:{\color{blue}440}$ committees with 3 women, 1 man (with Jill).

(b) Jill is not on the committee.
Choose 3 women from the other 11 women: .$\displaystyle {11\choose3} \:=\:165$ ways.
Choose 1 man from the available 9 men: .$\displaystyle {9\choose1}\:=\:9$ ways.

Hence, there are: .$\displaystyle 165\cdot9 \:=\:{\color{blue}1485}$ committees with 3 women, 1 man (without Jill).

Therefore, there are: .$\displaystyle 495 + 440 + 1485 \:=\:\boxed{2420}$ committees
. . which have more women than men, and Jack and Jill do not serve together.