(Counting Techniques) Why doesn't this approach work?

[Chetlin]

It may just be because I probably got only an hour of sleep last night and my thinking skills are not very good, but I'm grading a question on a final exam in discrete math and a lot of stiudents gave an answer that is incorrect, but I can't see why it is incorrect.

Homework Statement

Given a group of 15 men and 12 women, how many ways can we form a committee of 5 people that includes at least 1 woman?

Homework Equations

not applicable here (?)

The Attempt at a Solution

The correct solution is to subtract the number of committees that contain only men from the total number of possible committees. There are ${15 \choose 5}$ committees that consist of only men and ${15 + 12 \choose 5} = {27 \choose 5}$ total possible committees, so the number of committees that include at least one woman is ${27 \choose 5} - {15 \choose 5} = 77,\!727$.

An approach that a lot of students made and I can't find fault with is to say that if you choose one woman for the first spot, then the rest of the spots in the committee can be anyone. So you have 12 choices for the first spot, 26 choices for the second spot, 25 choices for the third spot, 24 choices for the fourth spot, and 23 choices for the fifth spot. If you multiply these numbers together and then divide the product by 5! (since order does not matter and there are 5! ways that the committee members could be arranged), you get the result $\frac{12 \cdot 26 \cdot 25 \cdot 24 \cdot 23}{5!} = 35,\!880$ which is different from the correct solution. So there is a mistake here.

Another way to get the correct answer is to calculate the sum ${12 \choose 1}{15 \choose 4} + {12 \choose 2}{15 \choose 3} + {12 \choose 3}{15 \choose 2} + {12 \choose 4}{15 \choose 1} + {12 \choose 5}{15 \choose 0}$. This does give the correct result of 77,727 and it, in a way, seems like the "correct" way to use the approach (as opposed to the incorrect way that I talked about before that gave the incorrect answer).

Thanks a lot!

 

[Simon Bridge]

Use smaller numbers to reality check -
Committee of 2, choosing between three women and three men.
The women are Beth, Cath, and Doris.
The men are Andy, Eddie, and Ian.

Basically - you undercounted the number of permutations by counting only those permutations with a woman in the first slot. i.e. if there was a man in the first slot, you did not count that committee.

 

[Chetlin]

All right, thanks! I thought it was something like that but due to my sleep deprivation I couldn't formulate the complete thoughts required to understand it all the way.