Ok. I think I've got this. For n balls distributed randomly in N bins, the probability of picking a bin at random and finding x balls in it is:
C(n-x+N-2, N-2)/C(n+N-1, N-1)
which is the same as
(n-x+N-2)!(N-1)!n!/[(N-2)!(n-x)!(n+N-1)!]
for 50 balls distributed randomly between 20 bins the...
It's interesting to note that for 1 bin there is only one way. All the balls go in the one bin. For two cups there are k+1 ways (n=number of balls, k = number of cups) taking 7 balls for some examples:
70
61
51
41
31
21
11
n+1 ways
for 3 cups it is the sum of the natural numbers from 1 to...
I just wrote a long reply which I lost (that's what comes with being a noob). I have a similar answer to you but slightly different:
C((n+k-1),(k-1))
which is (n+k-1)!/(k-1)!(n-2)!
This gives 10 for putting 3 balls into 3 cups which is what I mentioned above. Here are the 10 ways:
300
210
201...
SW VandeCarr: I like your way of doing this but am having some problems understanding it fully. You seem to suggest a normal distribution around a peak 2.5. I have a problem with this because if you imagine a 3 ball deviation from the mean then you get a problem. Of course it is possible to have...
I believe solving this problem is what you do to derive the Maxwell-Boltzman distribution of energies in molecules at equilibrium, although in that case you have a very large number of bins (molecules) and balls (energy). That might be a good starting point for researching for a combinatorial...