Question: $4$ people are to be chosen from $6$ married couples such that there is exactly one married couple in the group.
For this question I first chose one married couple out of the $6$ : $\binom{6}{1}$.
Then the number of people left are 10. Now choosing one from these: $\binom{10}{1}$.
Now I can choose from $8$ other people (as one of them will be married to the one chosen above):$\binom{8}{1}$
So ultimately the number of ways is: $\binom{6}{1}\cdot\binom{10}{1}\cdot\binom{8}{1} = 480$.
But the answer is $240$. Where is my logic flawed?
Any help would be appreciated!
Best Answer
The mistake is in the last two steps. Imagine you first selected the first partner from the first couple (arbitrary choice) out of ten people, and then the second partner from the second couple out of eight people. The result would be the same as when you would first have chosen the second partner from the second couple, and then the first partner from the first. Thus, you must divide by $2$ to account for this.