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?
 A: You’re off on the wrong foot right away with that $6$P$6$ for the case in which the team is made up entirely of singles: there is only one such team, and you’re counting $6!=720$. We’re not picking six people and assigning each of them a specific rôle; we’re just picking a group of $6$ people. If we pick them from the singles, we can do it in $\binom66=1$ way.
I think that I’d break it down according to how many singles we choose. 


*

*As we just saw, there is one team consisting of $6$ singles. 

*To form a team with $5$ singles, we can pick the singles in $\binom65=6$ ways and the sixth person in $\binom{10}1=10$ ways for a total of $6\cdot10=60$ teams.

*To form a team with $4$ singles, we can pick the singles in $\binom64=15$ ways. And since we’re allowed one couple, we can pick any two of the other ten people, so there are $\binom{10}2=45$ to pick the other two members of the team. That gives us another $15\cdot45=675$ teams.

*$3$ singles can be picked in $\binom63=20$ ways, and we can pick any three of the other ten people, so we get another $20\binom{10}3=20\cdot120=2400$ teams.
Now it gets a little trickier, since we start running into restrictions on whom we can choose from the couples.


*

*$2$ singles can be picked in $\binom62=15$ ways. There are $\binom{10}4=210$ ways to pick $4$ people from the couples, but some of these are forbidden: specifically, we may not pick two couples. Since there are $5$ couples altogether, there are $\binom52=10$ pairs of couples. That’s $10$ sets of four people that we aren’t allowed to choose, leaving $210-10=200$ sets that we are allowed to choose. Thus, we can form $15\cdot200=3000$ teams with $2$ singles.

*One single can be picked in $6$ ways. There are $\binom{10}5=252$ ways to choose $5$ people from the couples, but here again some are not allowed. Specifically, we have to throw out those groups that consist of two-and-a-half couples. As before, there are $\binom52=10$ ways to pick two couples, and there are then $6$ ways to pick one more person from the remaining three couples. That makes $10\cdot6=60$ ways to fill out the team and gives us $6\cdot60=360$ teams.
There’s one more case, the teams containing no singles; I’ll let you have a chance to work it out on your own, but feel free to ask if you get stuck.
A: 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.
$$\sum_{i=0}^5 \binom{6}{6-i}\binom{5}{i}2^i +\binom{5}{1}\sum_{i=0}^4 \binom{6}{4-i}\binom{5}{i}2^i.$$
