The number of ways to put $n$ unlabeled balls in $k$ distinct bins is $$\binom{n+k-1}{k-1} .$$ Which makes sense to me, but what I can't figure out is how to modify this formula if each bucket has a max of $m$ balls.

EDIT: What I've tried:

I got to the generating function $$(1-x^{m+1})^k(1-x)^{-k}$$ which ends up giving me $$\sum_{r(m+1)+r_2=n} \binom{k}{r}(-1)^{r_2}\binom{k+r_2-1}{r_2}$$

But when programing this:

def distribute_max(total,buckets,mmax):
  ret = 0
  for r in xrange(total//(mmax+1)+1):
    r_2 = total - r*(mmax+1)
    ret += choose(buckets,r) * (-1)**r_2 * choose(buckets + r_2 - 1,r_2)
  return ret

I'm getting terribly wrong answers. Not sure which step I screwed up.

  • $\begingroup$ Have you tried it for $M=1$ and $M=2$? $\endgroup$ – Phira Dec 7 '11 at 23:11
  • $\begingroup$ @mayhewsw the link appears to be broken. $\endgroup$ – naktinis Oct 1 '18 at 12:04
  • $\begingroup$ Thanks @naktinis. Fixed link: complete derivation here. $\endgroup$ – mayhewsw Oct 2 '18 at 13:22

As a check I did it with an inclusion-exclusion argument, getting $$\sum_i(-1)^i\binom{k}i\binom{n+k-1-i(m+1)}{k-1}\,.$$

Setting $r=i$ and $r_2=n-i(m+1)$ to match your notation, I make this $$\sum_r (-1)^r\binom{k}r\binom{k+r_2-1}{r_2}\,.$$ It appears that you’ve the wrong exponent on $-1$.

  • 2
    $\begingroup$ +1. Charalambides's Enumerative Combinatorics (Exercise 9.6, p. 360) gives the same expression. $\endgroup$ – Mike Spivey Dec 7 '11 at 23:39
  • 3
    $\begingroup$ @Brian What are the sum limits in your formula? The expression from book Enumerative Combinatorics (p. $360$) is $L(n,k,m)=\sum_{j=0}^k (-1)^j C_k^j C_{k+n-j(m+1)-1}^{k-1}$ according to your notation, where $C_k^j$ means the binomial coefficient. Unfortunately, $L(n,k,m)=0$ for any $n \leq k \cdot m$ (Mathematica' showed me that). Somewhere the mistake lives. Can you kindly help me with this problem? $\endgroup$ – Piotr Semenov Feb 3 '14 at 21:20
  • 2
    $\begingroup$ See also Balls In Bins With Limited Capacities. $\endgroup$ – joriki Oct 3 '15 at 7:10
  • 1
    $\begingroup$ It seems to me that in order to make this work without failing one should have $0\le k\le \frac n{m+1}$. $\endgroup$ – Miles Eagle Mar 6 '19 at 9:55

Let's denote the problem as $D(n,k,m)$, then when $mk-n \leq m$, the answer can be given as: $D(n,k,m)=\binom{km-n+k-1}{k-1}$, where $m \leq n$.

It can be easily verified using $D(5,2,3)=2$, or $D(6,3,3)=10$, etc..

Let me try to explain it in more details with my poor English (sorry for that).

Suppose we start from the state that all bins are filled with $m$ balls in each. Then the task for us is to eliminate $mk-n$ balls from these $k$ bins. To ditribute the $mk-n$ "elimination" into $k$ bins, we have $\binom{km-n+k-1}{k-1}$ different ways if $mk-n \leq m$, i.e. the number of elimination in each bin will not exceed $m$.

  • $\begingroup$ Why don't you write down the solution more explicitly? $\endgroup$ – Babai Mar 16 '15 at 8:35

The number of ways to put $n$ balls in $k$ boxes with in each box a maximum of $m$ balls is the coefficient of $[q^n]$ in the $\text{QBinomial}(k+m,k,q)$.

For example, the number of different ways to put $n = 5$ balls in $k = 4$ boxes with in each box no more than $m = 3$ balls is given by
$$ \text{QBinomial}(7,4,q) = (q^6+q^5+q^4+q^3+q^2+q+1)(q^2-q+1)(q^4+q^3+q^2+q+1) = q^{12}+q^{11}+2q^{10}+3q^9+4q^8+4q^7+5q^6+4q^5+4q^4+3q^3+2q^2+q+1. $$ The coefficient of $q^5$ is $4$. Hence there are four ways. These four ways are $(2,3)$, $(1,1,3)$, $(1,2,2)$, $(1,1,1,2)$.

  • $\begingroup$ The boxes are distinct, so there are 40 ways for the n=5,k=4,m=3 example. In particular, 12 each for (2,3), (1,1,3) and (1,2,2), and 4 for (1,1,1,2). The code in the question, with the correction from the accepted answer, gives 40. $\endgroup$ – Evgeni Sergeev May 16 '17 at 4:50

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