Separating 18 people into 5 teams 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.]

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 $12/18 \cdot 3/17 + 6/18 \cdot 2/17 = 8/51$
