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A teacher wants to divide her class of 18 students into 5 teams to work on projects, with two teams of 3 students each and three teams of 4 students each.
a) In how many ways can she do this, if the teams are not numbered?
b) What is the probability that two of the students, Mia and Max, will be on the same team?
[This is not a homework problem.]
Assume first that the teams are labelled, with the two teams of three having labels A and B, and the three teams of four having labels C, D, and E.
There are $\binom{18}{3}$ ways to choose the people who will be on Team A. For each of these ways, there are $\binom{15}{3}$ ways to choose Team B. For every way to choose Teams A and B, there are $\binom{12}{4}$ ways to choose Team C, and then $\binom{8}{4}$ ways to choose Team D, and finally, if we like, $\binom{4}{4}$ ways to choose Team E.
However, when we remove the labels, the number of choices for the $2$ three-person teams gets divided by $2!$, and the number of choices for the $3$ four-person teams gets divided by $3!$, for a total of $$\frac{\binom{18}{3}\binom{15}{3}\binom{12}{4}\binom{8}{4}\binom{4}{4}}{2!3!}.$$
For the Mia and Max problem, we could count. But we prefer to work directly with probabilities.
Imagine that the people first get divided into $2$ groups, of $6$ and $12$, to make up the two kinds of team.
The probability that Mia is selected for the group of $6$ is $\frac{6}{18}$. Given that this has happened, the probability Max is chosen for the same team is $\frac{2}{17}$. Thus the probability Mia and Max are in the same team of three is $$\frac{6}{18}\cdot\frac{2}{17}.$$ Similarly, the probability that Mia and Max are in the same group of $4$ is $$\frac{12}{18}\cdot\frac{3}{17}.$$ Add.
Remark: Doing problem (b) by counting and dividing is not difficult, just a little more messy-looking. Note that the probabilities are the same for the labelled teams case as for the unlabelled case. It is easier not to make a mistake by counting the number of ways that Mia and Max can be on the same labelled team, and dividing by the number of labelled teams.
The answer to the first question is just a multinomial coefficient divided by $2! \cdot 3!$ to make up for the fact that the teams are not numbered and we have therefore counted each partition into teams multiple times: $${\binom{18}{3, 3 ,4,4,4}}/(2!\cdot 3!) = \frac{18!}{(3!)^2 * (4!)^3 \cdot (2! \cdot 3!)}=1.072. 071. 000$$
The second question:
Probability that Max is on a 4-person team is $12/18$, probability that he is on a 3-person team is $6/18$. So the chance that mia is on the same team is $12/18 \cdot 3/17 + 6/18 \cdot 2/17 = 8/51$
Form each team one at a time.
$$\begin{align} \text{First team: }&\binom{18}{3}\\ \text{Second team: }&\binom{15}{3}\\ \text{Third team: }&\binom{12}{4}\\ \text{Fourth team: }&\binom{8}{4}\\ \text{Fifth team: }&\binom{4}{4} \end{align}$$
Then apply the rule of products.
The probability of any given student going into team one or two is $\frac{3}{18}$. The probability of a student going into team three, four, or five is $\frac{4}{18}$. So the probability of two students both going into team one is $\frac{3}{18}\frac{3}{18}$. Use the rule of sums to compute the total probability that they'll end up in the same team out of any of the five teams:
$$P=2\frac{3}{18}\frac{3}{18}+3\frac{4}{18}\frac{4}{18}$$
Starting with the "binomial(18, 3)+binomial(15, 3)+binomial(12, 4)+binomial(8, 4)+binomial(4, 4)" solution I got 1837 possible combinations, but if you start with dividing into the 4 teammember teams first I get "binomial(18, 4)+binomial(14, 4)+binomial(10, 4)+binomial(6, 3)+binomial(3, 3) = 4292 possible combinations", no idea if it's right or not just something to double check.
