Question

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

Answer 1

We break it into 2 parts

(1) Find out how many committees are possible with at least 2 from each party (there are 3 parties). This number will have no restrictions on majority.

(2) Find out how many possible cases are there to get a majority and subtract this value we get from (1).

Case 1: Out of 17 politicians we need at least 2 from each party. So there are 11 left. Therefore by the rule of grouping, the number of ways of grouping n identical things into r groups, where 0 is also possible = ^n+r-1C_r-1. Here n = 11, r = 3.

Therefore total possibile groups = ^11+3-1C_3-1 = ¹³C₂ = 78 groups (Where Majority is not a concern)

Case 2:Lets see How many possible ways we can get a majority. No Group should have 9 or more members, as that would constitute a Majority of all the members. Lets assume that the democrats get a majority and find the number of possibilities. This value will be multiplied by 3, as the same possibilities will arise for Republicans and Independents.

We have 11 left as 2 are already in each group. To get 9 or more Democrats, with 2 already being there, the following are the possibilities.

With 7 Democrats: (7,3,1) (7,1,3), (7,2,2) (7,0,4)(7,4,0) = 5

With 8 Democrats: (8,2,1), (8,1,2), (8,3,0) (8,0,3) = 4

With 9 Democrats: (9,1,1), (9,2,0), (9,0,2) = 3

With 10 Democrats: (10,1,0), (10,0,1) = 2

With 11 Democrats: (11,0,0) = 1

Total cases where Democrats have a Majority = 5 + 4 + 3 + 2 + 1 = 15

Therefore total cases, where there will be a majority = 15 * 3 = 45

therefore possible groups = 78 - 45 = 33 groups (Where these dont have a majority)