Suppose you have a group of 50 students, and you want to have two specific groups that have 10 and 15 students. What are the possible number of ways to arrange the two groups of students? What about the probability of getting student A and student B to NOT be in the same group?
I am confused on how you would take into account the rest of the students NOT inside the two groups mentioned in the problem. I assumed it would be something like
50 choose 10 * 40 choose 15 * 25 choose ?
Im not sure if my thought process is correct, and I would appreciate any help or hints.
Best Answer
You can pick $k$ out of $n$ students in $\binom{n}{k}$ ways. To pick two groups of 10 and 15 students out of a class of 50, you pick 10 out of 50 and then 15 out of the remaining 40, in all:
$\begin{align*} \binom{50}{10} \binom{40}{15} &= \frac{50!}{10! \cdot 40!} \cdot \frac{40!}{15! \cdot 25!} \\ &= \frac{50!}{10! \cdot 15! \cdot 25!} \\ &= \binom{50}{10, 15, 25} \end{align*}$
The last is a multinomial coefficient, the number of ways to divide a a class of 50 into groups of 10, 15 and 25.