Sometimes, you may make a combinatorial argument which seems reasonable, but you find that your answer is larger than the correct answer. Here is a problem which may trip you up: There are 30 liberals and 30 conservatives. How many ways can you form an 8-person committee if it must contain at least 3 liberals and at least 3 conservatives? The following is an incorrect solution: We choose 3 conservatives from 30, 3 liberals from 30, and 2 from the remaining 54 people, and our answer is {30\choose 3}^2 {54\choose 2}. Without reading further, think about why this is incorrect.

I will not explain why the solution above is incorrect; it is very important that you figure out why. The correct way of doing the problem is to consider each case separately. There are three cases, which are: choosing 3 liberals and 5 conservatives, 4 of each, and 5 liberals and 3 conservatives. So we see that the correct answer is
\displaystyle \sum_{k=3}^{5}{30 \choose k}{30 \choose 8-k} \displaystyle ={30 \choose 3}{30 \choose 5}+{30 \choose 4}{30 \choose 4}+{30 \choose 5}{30 \choose 3} \displaystyle =2{30\choose 5}{30 \choose 3}+{30 \choose 4}^2.

Solve the following problem (the solution will be posted in a week). There are 30 liberals and 30 conservatives. How many 8-person committees can be formed if at least one of them must be a liberal?