TITLE: Flaw in reasoning: number of ways to choose at least 1 man and 1 woman
QUESTION [9 upvotes]: Problem: A club has 5 men and 7 women. How many ways are there to choose 4 people such that at least one man and woman are chosen?
The correct answer is 455 and can be obtained by
$${12 \choose 4} - {5 \choose 4} - {7 \choose 4}.$$
However, I am unsure of the flaw in the following reasoning below which leads to an answer of 1575:
$${5 \choose 1}{7 \choose 1}{10 \choose 2}.$$
This quantity ensures that a man and woman are chosen, and then we select two of the remaining 10 people. As a result, we have chosen 4 people, of which there are at least 1 man and 1 woman.

REPLY [3 votes]: Hint: Suppose the men are selected from $\{\text{Adam, Bob, Carl, Dennis, Edgar}\}$ and the women from $\{\text{Alice, Barbara, Chloe, Darcy, Emily, Fiora, Gertrude}\}$.
We choose one man, say $\text{Bob}$ and one woman, say $\text{Alice}$. Then we choose two from ten that are neither of the already selected, say $\text{Emily}$ and $\text{Fiora}$.
Now for another group, choose one man, say $\text{Bob}$ once again. Then choose a woman, say $\text{Emily}.$ Now choose two out of the remaining ten, $\text{Alice} $ and $\text{Fiora}$.
Notice we've double counted the group of $\text{Alice, Bob, Emily, Fiora}$, using the latter method.