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I'm having trouble figuring out why these two different ways to write this combination give different answers. Here is the scenario:

Q: Choose a group of 10 people from 17 men and 15 women, in how many ways are at most 2 women chosen?

Solution A: From 17 men choose 8, and from 15 women choose 2. Or from 17 men choose 9, and from 15 women choose 1. Or from 17 men choose 10.

C(17,8)*C(15,2)+C(17,9)*C(15,1)+C(17,10) = 2936648 ways

Solution B: Choose from the men to fill the first 8 positions and choose the next 2 positions from the remaining men and women.

C(17,8)*[C(9,2)+C(9,1)*C(14,1)+C(14,2)] = 6150430 ways

What is wrong with my logic or interpretation here?

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The second method over counts. Specifically, if $X$ is one of the men, then you might choose him out of the first $8$ or you might choose him in the next two. – lulu Jan 10 at 0:31
The second way count some examples multiple times. – Thomas Andrews Jan 10 at 0:37
up vote 4 down vote accepted

Suppose that the men are $M_1,\ldots,M_{17}$, and the women are $W_1,\ldots,W_{15}$. Consider the group


Your second approach counts this $9$ times: once as $\{M_1,\ldots,M_8\}$ for the $8$ men and $\{M_9,W_1\}$ for the last two, once as $\{M_1,M_2,M_3,M_4,M_5,M_6,M_7,M_9\}$ for the $8$ men and $\{M_8,W_1\}$ for the last two, and so on. It does this overcounting for every group that contains exactly one woman.

Groups that contain no women are overcounted even more: for any group of $10$ men, there are $\binom{10}2=45$ different ways to split it into the first $8$ and the last $2$, so it gets counted $45$ times!

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Consider a simpler problem. How many ways are there to choose ten men from a set of ten men? By your second reasoning, choose eight first, then choose another two. That gives a total of: $$\binom{10}{8}\binom{2}{2}=45$$ Why is that reasoning problematic?

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