Creating a committee of 5 people with at least 3 men. Can't figure out why my reasoning is wrong. I'm having a hard time understanding why my thinking is wrong with this exercise:

From a group of 7 men and 6 women, five persons are to be selected to
  form a committee so that at least 3 men are there in the committee. In
  how many ways can it be done?

My reasoning is as follows:


*

*You need to a group of 3 men no matter what, so there are $7\choose 3$ = $35$ ways to pick this group. You now have 35 different groups of 3 men who can be combined with the other group of 2 people.

*After picking the 3 men, you can form a group of 2 people from the remaining 10 people in $10\choose 2$ = $45$ ways.

*The 3 men can be combined with the 2 other people in $35 * 45$ = $1575$ ways, which in my head should be the answer.


Now, I know this answer is wrong, and I know how to get to the real answer, but I can't understand what is wrong with this reasoning, because in my head it makes perfect sense.
So if you could please point out the fallacy in my thinking, that would be great!
Thank you in advance.
 A: Let the men be $m_1 \dots m_7$ and the women be $w_1 \dots w_6$.  How many times have you counted the committee  $\{m_1, m_2, m_3, m_4, m_5\}$?


*

*$\{m_1, m_2, m_3\}$ in your step 1 and $\{m_4, m_5\}$ in your step 2, 

*$\{m_1, m_2, m_4\}$ in your step 1 and $\{m_3, m_5\}$ in your step 2, 

*$\{m_1, m_3, m_4\}$ in your step 1 and $\{m_2, m_5\}$ in your step 2, 

*$\{m_2, m_3, m_4\}$ in your step 1 and $\{m_1, m_5\}$ in your step 2, 

*et c...


You've overcounted.
A: You are counting each committee with more than three men multiple times.
The number of ways of selecting a committee with exactly $k$ of the seven men and $5 - k$ of the six women is 
$$\binom{7}{k}\binom{6}{5 - k}$$
Since there are at least three men, the number of five-person committees with at least three men is 
$$\binom{7}{3}\binom{6}{2} + \binom{7}{4}\binom{6}{1} + \binom{7}{5}\binom{6}{0}$$
You count each committee with four men four times, once for each way of designating three of those men as the three men on the committee.  Say the men on the committee are Andrew, Bruce, Charles, and David and the woman is Elisabeth.  You count the committee in each of the following ways:
\begin{array}{l & l}
\text{three men} & \text{additional people}\\
\text{Andrew, Bruce, Charles} & \text{David, Elizabeth}\\
\text{Andrew, Bruce, David} & \text{Charles, Elizabeth}\\
\text{Andrew, Charles, David} & \text{Bruce, Elizabeth}\\
\text{Bruce, Charles, David} & \text{Andrew, Elizabeth}
\end{array}
You count each committee with five men ten times, once for each way you can designate three of them as being the three men on the committee.
Notice that 
$$\binom{7}{3}\binom{6}{2} + \color{red}{\binom{4}{3}}\binom{7}{4}\binom{6}{1} + \color{red}{\binom{5}{3}}\binom{7}{5}\binom{6}{0} = \color{red}{\binom{7}{3}\binom{10}{2}}$$
The error arises since your set of seven men from which the three men are drawn is not disjoint from your set of ten additional people from which the other two people are drawn.
