In how many ways can 12 golf balls be distributed to 4 golfers in each of the following cases?
b. The balls are different
3 balls each.
so suppose we give them out in order
we have \binom{12}{3} possibilities of balls for the first guy
then \binom{9}{3} possibilities of balls for the second guy
then \binom{6}{3} for the third
and whatever is left over for the last guy
this gives$\displaystyle \binom{12}{3} \cdot \binom{9}{3} \cdot \binom{6}{3}$ ways
but we have not acounted for the order in which we distributed the balls so each distrubtion of the balls is repeated $\displaystyle 4!$ times.
giving a final answer of $\displaystyle \frac{\binom{12}{3} \cdot \binom{9}{3} \cdot \binom{6}{3}}{4!} = 15400 $
If the instructions say that some of the golfers may receive no balls, then galactus has given you the answer.
However, if we require that each golfer receives at least one ball the we are counting surjections: $\displaystyle \mbox{surj}(12,4) = \sum\limits_{j = 0}^4 {( - 1)^j {4 \choose j}(4 - j)^{12} } $.
And what would be the solution if the balls are indistinctive?
Namely, I have a similar problem. In how many ways can one put n blank sheets in k drawers, whereas a drawer can be empty?
e.g. 100 sheets, 3 drawers
d1 d2 d3
100 0 0
0 100 0
23 45 32
etc.
Thanks!