0
$\begingroup$

There are 11 people who want to participate in a quiz. The maximum size of a team is 3, but no one from the 11 people want to compete alone. In how many different groupings can they enter the competition?

I'm struggling to get the different groupings because there seems to be a large amount of possibilities: first the number of people in a group, then you would have to consider who's in each of them. Is there an equation to solve this? If not, is there an easier way instead of just counting all the possibilities?

  • 0
    Well, straight counting isn't that bad. I'd start by listing the patterns: Three teams of three and one team of two, and so on. There aren't very many. Then fix a pattern and count the number of ways to populate the teams. – 2017-02-04

1 Answers 1

1

A little thought will show that there must be an odd number of triples,
so in fact only two patterns are possible:$\;3-3-3-2\;$ or $\;3-2-2-2-2$

and distribution into groups will be $\dfrac{\binom{11}3\binom83\binom53\binom22}{3!}\;$ + $\; \dfrac{\binom{11}3\binom82\binom62\binom42\binom22}{4!},$

The divisors $3!$ and $4!$ are needed as the groups would seem to be unlabelled.