Class divided into 5, probability of 2 people in the same team This might be a very simple question, but: a class of 25 students is divided into 5 teams of 5 each. What is the probability of student X and Y being in the same team?
is it just 4/25?
 A: Imagine that our heroes, A and B, are assigned to teams, in that order, with the rest being assigned later.
Whatever team A is assigned to, there are $4$ empty spots on that team. The probability B is given one of these spots is $\frac{4}{24}$. 
Or else we can do more elaborate counting. Imagine the teams are labelled (it makes no difference to the probability). There are 
$\binom{25}{5}\binom{20}{5}\binom{15}{5}\binom{10}{5}\binom{5}{5}$ ways to assign the people to labelled teams, all equally likely.
Now we count the number of ways A and B can end up on the same team. Which team? It can be chosen in $5$ ways. The other $3$ people on that team can be chosen in $\binom{23}{3}$ ways. And the rest of the assignments can be done in $\binom{20}{5}\binom{15}{5}\binom{10}{5}\binom{5}{5}$ ways. 
Divide. We get that the probability is $\frac{5\binom{23}{3}}{\binom{25}{5}}$. This simplifies to $\frac{1}{6}$. 
A: Hint.  Once you know which team X is on, how many places altogether are available for Y?  And how many of those available places are in the same team as X?
A: Another way to look at this is to think of how many ways you can arrange 25 kids into five five-member teams.  This is $\frac{25!}{5!5!5!5!5!}$. 
Now, how many ways can you arrange it so that $X$ and $Y$ are on the same team?  In whatever team they end up (so there are 5 such equivalent scenarios) the other 23 kids will be arranged around them, 3 kids on the team our buddies ended up in, while the others will be spread out in teams of five (i.e., 3/5/5/5/5 or 5/3/5/5/5, or 5/5/3/5/5, or ... , you get the idea).  
Each of these scenarios has probability $\frac{23!}{3!5!5!5!5!5!}$. Thus, the final probability will be
$$
P(X, Y on\,\,same\,\,team)=\frac{\frac{23!}{3!5!5!5!5!5!}}{\frac{25!}{5!5!5!5!5!}}\times 5=\frac{23!5}{25!3!}\times 5=\left (\frac{5}{25}\right )\left (\frac{4}{24}\right )5=\left (\frac{4}{24}\right )=\frac{1}{6}
$$
