Number of ways to put $n$ unlabeled balls in $k$ bins with a max of $m$ balls in each bin

Question

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.

Answer 1

As a check I did it with an inclusion-exclusion argument, getting

$$\sum_{i=0}^{\left\lfloor\frac{n}{m+1}\right\rfloor}(-1)^i\binom{k}i\binom{n+k-1-i(m+1)}{k-1}\,.\tag{1}$$

The nature of the inclusion-exclusion argument is such that $\binom{k}i\binom{n+k-1-i(m+1)}{k-1}$ is the number of ways to distribute the $n$ balls so that at least $i$ of the $k$ bins have more than $m$ balls. In that case some $i$ bins contain altogether at least $i(m+1)$ balls, so clearly we must have $i(m+1)\le n$, i.e., $i\le\left\lfloor\frac{n}{m+1}\right\rfloor$ (since $i$ is an integer).

Clearly it is also the case that $i$ cannot exceed $k$, but we don’t take explicit note of that in the upper limit of the summation $(1)$, because any term with $i>k$ has a factor $\binom{k}i=0$ anyway. One could make this explicit by writing the upper limit as $\min\left\{k,\left\lfloor\frac{n}{m+1}\right\rfloor\right\}$, but it is not necessary to do so.

Setting $r=i$ and $r_2=n-i(m+1)$ to match your notation, I make this

$$\sum_{r=0}^{\left\lfloor\frac{n}{m+1}\right\rfloor}(-1)^r\binom{k}r\binom{k+r_2-1}{r_2}\,.$$

It appears that you’ve the wrong exponent on $-1$.

Answer 2

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 $m$.

Answer 3

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 $(2,3)$, $(1,1,3)$, $(1,2,2)$, $(1,1,1,2)$.