# Combinations: dividing students into 3 groups [duplicate]

How would you solve a problem: we have $$30$$ students in a classroom. How could we divide them into three groups, if there needs to be $$5$$ students in the first group, $$10$$ students in the second group and 15 students in the third group. The result is: $$\frac{30!}{5!×10!×15!}$$ . Why isnt it $$\frac{30!}{3!×5!×10!×15!}$$ since we have three groups with $$3!$$ permutations.

• Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Oct 6 '19 at 8:00
• $\binom{30}{5}\binom{25}{10}\binom{15}{15}$ is the correct answer – user655800 Oct 6 '19 at 8:17
• But that result is written in the book and our teacher solved it on the board, but I still dont get it... how to come to that result – jdmath Oct 6 '19 at 8:32
• The groups are distinguished by their sizes. Choosing who is in each group completely determines the groups. If you had groups with the same size, then you would have to divide by the number of ways you could choose the same groups. – N. F. Taussig Oct 6 '19 at 9:14
• I have addressed your question here. – N. F. Taussig Oct 6 '19 at 10:38