Why is the following method incorrect when deciding on the possible choices? "Six men and three women are standing in a supermarket queue. Three of the people in the queue are chosen to take part in a customer survey. How many different choices are possible if at least one woman must be included?"
I went about solving this question by considering 3 cases, first with a single woman and then with 2 and finally with 3.
$$
 {3 \choose 1} \ \times \ {6 \choose 2} \ + \  {3 \choose 2} \ \times \ {6 \choose 1}  \ + \  {3 \choose 3}
$$
Which simplifies to $ \ 45 \ + \ 18 \ + \ 1$ leading to $64$ different choices.
This approach is in fact correct. However, another approach came to my mind as well.
If at least 1 woman must be included in the group then we can simply choose 1 from the 3 women, and fill the remaining 2 slots from the remnant 8 'people'.
$$
{3 \choose 1} \ \times \ {8 \choose 2}
$$
This, however, gives the answer $84$. This answer is most certainly wrong but I am unable to explain why the method is incorrect. If someone could explain why this leads to the wrong answer that would be very nice.
 A: You are counting each selection with two women twice and each selection with three women three times, once for each way you could have designated one of the women who is selected as the woman you have selected and the remaining woman or women as additional people.
Suppose you select two women, Abigail and Beatrice, and one man, Charles.  You count this selection twice:
\begin{array}{l l}
\text{woman} & \text{additional people}\\ \hline
\text{Abigail} & \text{Beatrice, Charles}\\
\text{Beatrice} & \text{Abigail, Charles} 
\end{array}
Suppose you select three women, Abigail, Beatrice, and Charlotte.  You count this selection three times:
\begin{array}{l l}
\text{woman} & \text{additional people}\\ \hline
\text{Abigail} & \text{Beatrice, Charlotte}\\
\text{Beatrice} & \text{Abigail, Charlotte}\\ 
\text{Charlotte} & \text{Abigail, Beatrice} 
\end{array}
Notice that
$$\binom{3}{1}\binom{6}{2} + \color{red}{\binom{2}{1}}\binom{3}{2}\binom{6}{1} + \color{red}{\binom{3}{1}}\binom{3}{3} = 84$$
where the factors in red are the number of ways you could have designated one of the women you have selected as the woman you have selected.
A: There is a much simpler approach. "At least one woman" is the opposite of "zero women" or "all men" from the total, so subtracting from the total gives us:
$${9 \choose 3} - {6 \choose 3} = 64$$

If you are interested in another subtractive approach, following from N. F. Taussig's answer, you can again subtract from ${3 \choose 1} \cdot {8 \choose 2} = 84$.
The cases where there are two women and one man (Abigail, Beatrice, Charles) are counted $2$ times, when they should only be counted $1$ time. Therefore, you need to subtract the number of combinations of this type $1$ time. This is just ${3 \choose 2} \cdot 6 = 18$.
Similarly, the cases where there are three women are counted $3$ times, when they should only be counted $1$ time, as there is only one combination possible with all $3$ women. Therefore, the number of combinations should be subtracted by $2$.
Hence $84 - 18 - 2 = 64$, which is the correct answer.
A: Because you can choose Becky with ${3 \choose 1}$ and Kate with ${8 \choose 2}$, and in another way choose Kate with ${3 \choose 1}$ and Becky with ${8 \choose 2}$ all other choices being the same. You repeat cases depending on which order the women are chosen if there is more than $1$.
