Taking 6 people from 8 married couples where there is exactly one married couple 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

*

*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.


*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.


*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?
 A: We need to select one married couple, which, as you say, can be done in $\binom{8}{1}$ ways.  We also need to choose four more people, with one each from four of the remaining seven married couples.  Choose which four of those seven couples will have a representative in $\binom{7}{4}$ ways.  For each such couple, there are two ways to select the representative from that couple.  Hence, there are $$\binom{8}{1}\binom{7}{4}2^4 = 4480$$
ways to select six people from eight married couples such that there is exactly one married couple among those six people.
A: 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 $0.1$
A: Choose $5$ soloists from the $8$ couples: $\binom{8}{5}2^5=56\cdot32=1,792$.
Choose $1$ from the $5$ soloists to be re-united: $\frac52$ as the soloist never was.
Answer: $4,480$.
(or, 5 couples from the 8, then 'the' couple and 4 soloists)
Your method is wrong as you have double-counted.
