Combinations problem with a group and 2 subgroups From 50 school students we have to choose 6 students that will dance in a school play and 4 students that will be the actors in the play.
How many different combinations of dancers and actors can we make?
I wrote $\dfrac{50!}{6!4!}$. Am I right? Or is it $C(50,6)·C(44,4)$?
It's something that confuses ME.
Thank you very much in advance.
 A: You're right that it's ${50 \choose 6}{44 \choose 4}$: you first choose $6$ students from among $50$, then $4$ from the remaining $44$. Notice that you could alternatively choose the actors first, then the dancers: you would have ${50 \choose 4}{46 \choose 6}$, which gives the same result. (Verify this!)
Look at what happens when you expand:
$${50 \choose 6}{44 \choose 4} = \frac{50!}{6!44!}\frac{44!}{4!40!} = \frac{50!}{6!4!40!}.$$
You can interpret this as follows: look at all the ways of ordering $50$ people (there are $50!$ such ways). Put them in groups of $6$, $4$, and the remaining $40$. But we don't care about the order within those groups, so we divide by $6!$, $4!$, and $40!$.
Coming back to your original guess, $\frac{50!}{6!4!}$, you can see that the only difference is that you didn't divide by $40!$. This isn't right, but it still has a meaning: namely, you do care about the order of the remaining $40$ people. For example, this would be the right answer if the situation were: "Out of $50$ students, we need to choose $6$ to be dancers, $4$ to be actors, and the remaining $40$ will line up to get tickets (and we care about the order of the line)."
