Different answers from different formulations of combinatorics problem

Question

A certain men's club has sixty members; thirty are business men and thirty are professors. In how many ways can a committee of eight be selected if at least three must be business men and at least three professors?

I figured that I could do this two ways. Either I could deal with the cases individually and get: $C(30,3)\times C(30,5) \times 2 + C(30,4)^2$, or I thought I could choose three business men, choose three professors, and then choose 2 from the remaining 54: $C(30,3)^2 \times C(54,2)$ like so.

Unfortunately, these yield two very different answers. What am I missing? I'm sure it's dumb.

Answer 1

Your first calculation is correct. The second overcounts. To see why, consider the committee $\{P_1,P_2,P_3,P_4,B_1,B_2,B_3,B_4\}$, where the $P_k$ are professors and the $B_k$ are businessmen. This committee gets counted once with $P_1,P_2$, and $P_3$ as the three professors initially chosen and $B_1,B_2$, and $B_3$ as the three businessmen initially chosen, $\{P_4,B_4\}$ being the odd pair. But it’s also counted with, e.g., $P_2,P_3$, and $P_4$ as the three professors initially chosen, $B_1,B_3$, and $B_4$ as the three businessmen initially chosen, and $\{P_1,B_2\}$ as the odd pair. It might be instructive to try to count just how many times this one committee is counted by your second calculation.

Answer 2

The first method is correct.

When you compute the answer as $C(30,3)^2 \times C(54,2)$, you are treating the last two chosen members as "distinguished", when they aren't, and so you are overcounting.

Answer 3

On the "teach a man to fish" principle, let me show you how you could have answered your own question. Could be useful on an exam.

If the numbers weren't so damn big, you could list all the committees you get by each method, and then examine them to see which method was missing some instances, or double counting, or counting some erroneous solutions. Instead, you can:

Try a similar problem with smaller numbers to see what's going on.

The smallest numbers that don't trivialize the problem seem to be: $3$ businessmen, $3$ professors, committees of $3$ containing at least one of each. Doing it by your second method, you get:$$\binom31^2\binom41=36.$$Obviously wrong, because all told there are only $\binom63=20$ ways to pick a committee of $3$. Now you can list the $36$ committees you counted and see where you overcounted. Calling the businessmen a,b,c and the professors x,y,z:

axb AXC axy axz ayb ayc ayx ayz azb azc azx azy
bxa bxc bxy bxz bya byc byx byz bza bzc bzx bzy
CXA cxb cxy cxz cya cyb cyx cyz cza czb czx czy