Q:
(a) A committee of 3 is to be chosen from a group of 9 politicians.
Suppose two of the group belong to Party I, three of the group belong
to Party II and four of the group belong to Party III. How many of the
committees have at least two members from the same party? Explain your
answer.
Attempt:
Firstly I did 9C3 to get 84. This is just to get things started. I then did 2C1 for party 1, 3C2 for party 2 and 4C3 for party 3 and my answers were 2,3,4. I feel like this is wrong though.
Best Answer
The committee will have at least two members from the same party unless each member of the committee is drawn from a different party. Therefore, we can find the answer by subtracting the number of committees in which each person is drawn from a different party from the total number of three-person committees which can be drawn from the nine politicians.
As you found, there are $$\binom{9}{3}$$ ways to form a three-person committee drawn from the nine politicians.
The number of ways of obtaining a committee in which each person is drawn from a different party is $$\binom{2}{1}\binom{3}{1}\binom{4}{1}$$ since one of the two members of Party I, one of the three members of Party II, and one of the four members of Party III must be selected.
Therefore, the number of three-person committees which have at least two members of the same party is $$\binom{9}{3} - \binom{2}{1}\binom{3}{1}\binom{4}{1}$$