TITLE: How many options are there for 15 student to divide into 3 equal sized groups?
QUESTION [13 upvotes]: How many options are there for 15 students to divide into 3 equal-sized groups?
Now I know the solution is $\;\dfrac {15!}{5!5!5!3!}\;$ but I can't understand why.
Can anyone please enlighten me?

REPLY [3 votes]: I actually found a nice way of solving it (After reading the post on the comment):
Say i choose a person first in the line-up, Now that person have $\binom {14}4$ ways of choosing 4 other groups members, Then the next person in the line up has $\binom 94$ ways of choosing other groups members. Finally, the last member has $\binom 44=1$ ways of choosing his group members.
If we multiply all of the above: $\binom {14}4\times$$\binom 94$ and we get the exact same  result as stated. (Just written a bit differently.)