n indistinguishable balls into k distinguishable bins

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.

• I am confused: Don't you have $4$ bins in the first sentence? – zoli Dec 18 '16 at 10:57
• @zoli There are 3 bins, the top bin have at least 4 balls – blz Dec 18 '16 at 10:59
• Is there any problem with my answer? – zoli Dec 21 '16 at 22:31

$$\binom{7}{2},$$
because we have $5+2$ abstract objects ($2$ walls between bins and $5$ balls) and we have to select two of the $7$ objects to play the role of the walls.
So, to answer your question explicitly: Three bins are separated by $2$ (and not $3$) walls [floors]. $\color{red}{\text{That is, we are choosing two separators and not three bins.}}$