I have read almost identical problem in "Taking 4 people from 6 married couples with exactly one married couple selected" but I don't quite understand.
I modify the problem by increasing the taken people and available married couples as follows.
How many way are there to take 6 people from 8 married couples such that there is exactly only one married couple in the 6 people?
My attempt
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There are ${8\choose 1}= 8$ ways to choose the exactly one married couple. Two people are already selected. I need to take 4 people more.
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For every 2 married couples there are 4 possibilities to take 2 people who are not a married couple. So there are ${7\choose 2} \times 4$ ways to choose 2 more people. We still need 2 people more.
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With the same argumentation as the second point above, there are ${5\choose 2} \times 4$ ways to choose 2 remaining people.
As a result, there are ${8\choose 1}\times {7\choose 2} \times {5\choose 2} \times 4 \times 4 =26880$ ways.
However, could you tell me whether my answer is correct?
Best Answer
I think you are very close. I would say:
First we need to choose which married couple is included in its entirety. There are $8 \choose 1$ options for this $=8$ .
Now we will choose $4$ married couples from which we will just take one person. So we are picking $4$ couples from the remaining $7$. This is $7 \choose 4$ or $35$.
Finally, in each of the $4$ there are two ways of choosing which member of the couple we include so we multiply by $2^{4}=16$.
$8 \times 35 \times 16 = 4480$.
Do we believe this? Well let's do a sense check.
How many ways are there to choose $6$ people from $16$? $16 \choose 6$ = $8008$.
So the probability of getting one married couple is $4480/8008 = 0.56$ Yeah, I buy that, as in it is not over $1$ or under $0.1$