Help understanding combinatorics problem involving restrictions on committees

Question

How many ways are there to select a committee of $17$ politicians chosen from a room full of indistinguishable Democrats, indistinguishable republicans, and indistinguishable Independents if every party must have at least two members on the committee? If, in addition, no group may have a majority of the committee members?

I was able to find the number of ways to select a committee if each party must have at least two members on the committee: $\binom{11+2}{2}=\binom{13}{2}=78$ ways.

I am confused on the second part of the problem.

Any help is appreciated.

Answer 1

To have a majority means that one party has at least $9$ members. As it is impossible for two parties to each have $9$ members, you can compute the number of ways to form a committee with at least $9$ Democrats and at least two of each other party, multiply by $3$, and subtract from your $78$ ways without the restriction.

Answer 2

You have correctly calculated the total number of committees with at least two members from each party, $C(13,11)=C(13,2)$.

For the second part we can proceed as follows:

• In these cases the majority is defined as just more than half. Since half of 17 is 8.5, then the majority of the committee would be 9. We want to find the number of committees that present the majority in favor of a party.

• To guarantee that we will have a majority, we select 9 politicians from the same party. There are 3 options (parties) to choose from.

• We have to guarantee that the committees meet the condition of having at least two members from each party, so we select two members from the two remaining parties.

• So far we've filled $9+2+2=13$ slots. There are $17-13=4$ slots left that we can fill with politicians from any party. There are $C(3+4-1,4)=C(6,4)$ ways to select them.

• Finally we will have that there are $$3 \times C(6,4)$$ possible committees with a majority.

• We subtract this number from the total number of allowable committees you determined in part one: $$C(13,11)-3 \times C(6,4)=33$$ And we have obtained the number of admissible committees without a majority.