mixed permutations and combinations I have a problem that I am not too sure of. In a team of 16, there are 5 couples and 6 single people. In how many ways can at most 1 couple be chosen if 6 people are required to represent the team at a conference?
This is my solution: 6P6 (0 couples and 6 single people only) + 11C6 * 6! * 2! (choose 1 member of a couple and the other 6 single people, this can be done in 2!*6! ways) + 5C1 * 2! * 10C4 *2! * 4! (exactly one couple is chosen and any of the singles along with the other partners of the other couples)
Is this correct?

The following approach is reasonably systematic. There will be lots of words, but at the end there will be a more or less compact formula.
We count first the teams that have no couple, then, in basically the same way, the teams that have $1$ couple. A couple, viewed as an entity, will be called a family. There are $5$ families.
No couples: Maybe we choose $6$ singles. That can be done in $\binom{6}{6}$ ways. Of course this is $1$, but we call it by the complicated name $\binom{6}{6}\binom{5}{0}2^0$. Soon that will look reasonable!
Or else we pick $5$ singles, $1$ family, and a representative of the family. This can be done in $\binom{6}{5}\binom{5}{1}2^1$ ways.
Or else we pick $4$ singles, $2$ families, and a representative of each family. This can be done in $\binom{6}{4}\binom{5}{2}2^2$ ways.
Or else we pick $3$ singles, $3$ families, and a representative of each family. This can be done in $\binom{6}{3}\binom{5}{3}2^3$ ways.
Or else we pick $2$ singles, $4$ families, and a representative of each family. This can be done in $\binom{6}{2}\binom{5}{4}2^4$ ways.
Or else, finally, we pick $1$ singles, $5$ families, and a representative of each family. This can be done in $\binom{6}{1}\binom{5}{5}2^5$ ways.
Add up. A number of cases, but only one idea.
One couple: The idea is the same. There are $\binom{5}{1}$ ways o pick the couple. We will count the number of ways to pick the remaining $4$ people, add them up, and multiply by $\binom{5}{1}$. But rom here on, we only count the ways of picking the $4$.
We could pick $4$ singles. This can be done in $\binom{6}{4}$ ways, but we call the number $\binom{6}{4}\binom{5}{0}2^0$.
Or else we pick $3$ singles, $1$ family, and a representative of the family. This can be done in $\binom{6}{3}\binom{5}{1}2^1$ ways.
Or else we pick $2$ singles, $2$ families, and a representative of each family. This can be done in $\binom{6}{2}\binom{5}{2}2^2$ ways.
Or else we pick $1$ single, $3$ families, and a representative of each family. This can be done in $\binom{6}{1}\binom{5}{3}2^3$ ways.
Or else, finally, we pick $0$ singles, $4$ families, and a representative of each family. This can be done in $\binom{6}{0}\binom{5}{4}2^4$ ways.
Final answer: We gather the whole thing into a compact formula.
$$\sum_{i=0}^5 \binom{6}{6-i}\binom{5}{i}2^i +\binom{5}{1}\sum_{i=0}^4 \binom{6}{4-i}\binom{5}{i}2^i.$$
