my math teacher wrote a problem today:
In how many ways can you put 9 similar balls into 3 bins stacked on top of each other, so that the top bin will have at least 4 balls.
His answer was: let's put 4 balls into the top bin. then we're left with 5 balls into 3 bins, without order being important, and repetition (putting into the same bin) is allowed. Therefore the answer is $5+3-1 \choose 3-1$ = $7 \choose 2$ = 21
What I don't understand is why is it $5+3-1 \choose 3-1$ and not $5+3-1 \choose 3$? we're choosing 3 bins, not 2.