So the way to do this I think is to 6C3*7C3*9 (where 9 is the number of ways we can arrange the men and women within the 3 pairs).
But I can't seem to figure out how I would do this in a different approach where I look at the total pairs and pick 3 pairs out of it.
I initially thought, perhaps I can multiple 6C1 and 7C1 to get 42 pairs. From those 42, I can choose 3, 42C3. However, I notice there is an immediate problem since I have duplicates within this combination. How would I remove these duplicates?
Thanks in advance
Best Answer
Method 1: We correct your first approach.
There are only six ways of matching three men and three women in opposite sex pairs. Line up the three selected women in some order, say alphabetically. There are three ways to match the first woman in the line to one of the three selected men, two ways to match the second woman in the line to one of the two remaining men, and one way to match the last woman in the line with the remaining man. Hence, there are $3! = 6$ ways to match the three selected men with the three selected women.
We can verify this assertion by listing them. Let $M_1$, $M_2$, and $M_3$ denote the three selected men; let $W_1$, $W_2$, and $W_3$ denote the three selected women. The possible pairings are \begin{align*} & (M_1, W_1), (M_2, W_2), (M_3, W_3)\\ & (M_1, W_1), (M_2, W_3), (M_3, W_2)\\ & (M_1, W_2), (M_2, W_1), (M_3, W_3)\\ & (M_1, W_2), (M_2, W_3), (M_3, W_1)\\ & (M_1, W_3), (M_2, W_1), (M_3, W_2)\\ & (M_1, W_3), (M_2, W_2), (M_3, W_1) \end{align*}
Hence, a corrected version of your first approach is that there are $$\binom{6}{3}\binom{7}{3} \cdot 3!$$ ways of selecting three opposite sex pairs from six women and seven men.
Method 2: We choose the women, then match the men to the women.
There are $\binom{6}{3}$ ways of selecting three of the six women. We line them up in some order. We can match the first woman in the line to one of the seven men, the second woman in the line to one of the six remaining men, and the third woman in the line to one of the five remaining men, which yields $$\binom{6}{3} \cdot 7 \cdot 6 \cdot 5$$ ways of selecting three opposite sex couples.
Method 3: We choose three couples, then correct for the $3!$ orders in which the same couples could be selected.
There are seven ways to select a man and six ways to select a woman for the first pair, giving $7 \cdot 6 = 42$ ways to pair a man with a woman. That leaves six men and five women from which to choose the next pair, giving $6 \cdot 5 = 30$ ways to choose the second pair. Once those pairs have been selected, we are left with five men and four women from which to choose the third pair, giving $5 \cdot 4 = 20$ ways to choose the third pair. Thus, it would appear that the three pairs can be selected in $42 \cdot 30 \cdot 20$ ways. However, the same three pairs could be selected in $3!$ orders, as shown above. Hence, the number of ways of selecting three opposite sex couples is $$\frac{42 \cdot 30 \cdot 20}{3!}$$