Choosing at least 2 women from 7 men and 4 women In how many different ways can we choose six people, including at least two women, from a group made up of seven men and four women?
Attempt:
As we have to have at least two women in the choices, then $\displaystyle\binom{4}{2}$, leaving a total of $4$ out of $9$ people to be chosen. So $$\binom{9}{4} \cdot \binom{4}{2}=756$$
The answer is $371$. Where is my error?
 A: If you choose six people and at least two women are selected, then either two women and four men, three women and three men, or four women and two men are selected.  The number of ways of selecting $k$ women and $6 - k$ men from four women and seven men is
$$\binom{4}{k}\binom{7}{6 - k}$$
Hence, the number of selections with at least two women is
$$\sum_{k = 2}^{4} \binom{4}{k}\binom{7}{6 - k} = \binom{4}{2}\binom{7}{4} + \binom{4}{3}\binom{7}{3} + \binom{4}{4}\binom{7}{2} = 371$$
You counted each selection with more than two women multiple times.
Your method counts each selection with three women three times, once for each of the $\binom{3}{2}$ ways of designating two of the three selected women as the two women in the group, and each selection with all four women six times, once for each of the $\binom{4}{2}$ ways of designating two of the four women as the four women in the group.
For instance, suppose the women are Anne, Brenda, Charlotte, and Diana and the men are Edward, Frank, George, Henry, Ivan, Jeffrey, and Karl.  You count the selection of Anne, Brenda, Charlotte, Edward, Frank, and George three times.
$$\begin{array}{l | l}
\text{designated women} & \text{additional people}\\ \hline
\text{Anne, Brenda} & \text{Charlotte, Edward, Frank, George}\\
\text{Anne, Charlotte} & \text{Brenda, Edward, Frank, George}\\
\text{Brenda, Charlotte} & \text{Anne, Edward, Frank, George}
\end{array}
$$
Similarly, your method counts the selection of Anne, Brenda, Charlotte, Diana, Edward, and Frank six times.
$$\begin{array}{l | l}
\text{designated women} & \text{additional people}\\ \hline
\text{Anne, Brenda} & \text{Charlotte, Diana, Edward, Frank}\\
\text{Anne, Charlotte} & \text{Brenda, Diana, Edward, Frank}\\
\text{Anne, Diana} & \text{Brenda, Charlotte, Edward, Frank}\\
\text{Brenda, Charlotte} & \text{Anne, Diana, Edward, Frank}\\
\text{Brenda, Diana} & \text{Anne, Charlotte, Edward, Frank}\\
\text{Charlotte, Diana} & \text{Anne, Brenda, Edward, Frank}
\end{array}
$$
Notice that
$$\binom{4}{2}\binom{7}{4} + \color{red}{\binom{3}{2}}\binom{4}{3}\binom{7}{3} + \color{red}{\binom{4}{2}}\binom{4}{4}\binom{7}{2} = \color{red}{756}$$
A: First choosing $W_1$ and $W_2$ to satisfy the requirement, then choosing $W_3$ among the other four people gives the same group as first choosing $W_1$ and $W_3$, then choosing $W_2$ among the other four. Because of this, your method is counting many groups multiple times.
A: The number of ways is
$$\binom{4}{2}\binom{7}{4}+\binom{4}{3}\binom{7}{3}+\binom{4}{4}\binom{7}{2} =6\times35+4\times35+21 = 371$$
"At least two women" means either $2$, $3$ or $4$.
A: Consider three cases:

*

*choosing 2 women and 4 men

*choosing 3 women and 3 men

*choosing 4 women and 2 men

So the total number of arrangements is:
$\binom{4}{2} \times \binom{7}{4}+ \binom{4}{3} \times \binom{7}{3}+ \binom{4}{4} \times \binom{7}{2}=371$
