How many ways are there to place $l$ balls in $m$ boxes each of which has $n$ compartments?

Question

I asked another related question, but have not got any answer. So, I decided to break the question down into more simple questions. Here, you may see the main question.

Assume that there are $m$ boxes each of which contains $n$ compartments. There are also $l$ balls, where $\ m\ < \ l\ <\ (m-1)n+2$. Moreover, the balls are the same.

$\bullet$ I would like to know how many ways there are to place the $l$ balls into the $m$ boxes such that

$1-$ Each compartment can hold up to one ball.

$2-$ Each box must have at least one ball.

The probable solutions:

(a) If I first put a ball in each box, and then place the rest arbitrary through the compartments, the constraints will be satisfied. In this way, we first need to select a compartment of each box to put a ball. There are $\binom n1 ^m$ ways to do so. Then, to place the other balls arbitrary into the available compartments, there are $\binom{mn-m}{l-m}$ ways.

As a result, the answer will be $$C_1(m,n,l)\ =\ \binom n1 ^m \times \binom{mn-m}{l-m}$$

(b) Another answer may consider as $$C_2(m,n,l)\ =\ \sum_{i=0}^{k}(-1)^i \binom{m}{i}\binom{(m-i)n}{l}, $$ where $k = m - \lceil l/n\rceil $.

In this summation, the first term ($i=0$) counts the number of ways to place $l$ balls in $mn$ compartments with no constraint. The second term removes the cases that at least one of the boxes are empty. But, the second term, remove the cases with two empty boxes two times! So, in the third term, we add once such cases, and so forth.

$\bullet$ According to a simple example with $m=2,\ n=3,\ l=3$, we see that the answers are different. For this example, we have

$$C_1(2, 3, 3) = 3^2\times \binom 41 = 36,$$

$$C_2(2, 3, 3) = \binom 63 - \binom 21 \binom 33 = 20-2=18.$$

Please let me know your answers or at least your idea on the given answers. Why are they different?

Answer 1

To fix the context, let us clear that we are looking for the number of ways to arrange
- $l$ undistinguishable balls
- into $m$ distinguishable boxes, each provided with $n$ distinguishable compartments
- counting only the arrangements that contain no empty boxes.

As a visual help, let us reproduce an example with $m=2, \, n=3, \, l=3$.

Obviously, the total number of ways to put the $l=3$ undistinguishable balls (the "ones")
into the $m \cdot n = 6$ distinguishable places is $$ T(m,n,l) = \left( \begin{gathered} m\,n \\ l \\ \end{gathered} \right) = 20 $$ and deducting the first and last row, in which at least a box is totally empty, we get $18$ dispositions leaving no empty box, as correctly returned by the $C(m,n,l)$ formula.

That premised, there are various combinatorical and algebraic considerations to deduce, which renders this argument very interesting. I'll try and concisely expose the main ones.

In the formula for $C(m,n,l)$ the limits on the summation index $i$ may algebraically be omitted, as they are intrinsic in the two binomials when defined as $$ \left( \begin{gathered} x \\ q \\ \end{gathered} \right) = \left\{ {\begin{array}{*{20}c} {x^{\,\underline {\,q\,} } /q!} & {\left| {\;0 \leqslant \text{integer}\,q} \right.} \\ \text{0} & {\left| {\;\text{otherwise}} \right.} \\ \end{array} } \right. $$ which leaves more freedom in operating on the formula, while for computational purposes we can fix them to $0 \leqslant i \leqslant m$. So we have better write:

$$ C(m,n,l)\quad \left| {\;0 \leqslant \text{integer}\;m,n,l} \right.\quad = \sum\limits_{\left( {0\, \leqslant } \right)\;i\,\,\left( { \leqslant \,m} \right)} {\left( { - 1} \right)^{\,i} \left( \begin{gathered} m \\ i \\ \end{gathered} \right)\left( \begin{gathered} \left( {m - i} \right)n \\ l \\ \end{gathered} \right)} \tag {1} $$

Concerning the approaches in your post, note that the scheme leading to $C_1$ is over-shooting, because it makes distinction between the ones of "first " and "second" introduction within each box, and that is difficult to compensate. The Inclusion/Exclusion scheme is instead correct.

The example sketched indicates that the total number of arrangements can be partitioned into those that contain exactly $0$ empty boxes (${n}\choose {0}$ ways to choose that) and the associated $N(m,n,l)$ rows in the remaining boxes, plus exactly $1$ empty boxes (${n}\choose {1}$ ways) and the associated $N(m-1,n,l)$ rows, ...,
$k$ empty boxes associated to $N(m-k,n,l)$ rows, for $k$ ranging from $0$ to $m$, so that we can write

$$ T(m,n,l) = \left( \begin{gathered} n\,m \\ l \\ \end{gathered} \right) = \sum\limits_{\left( {0\, \leqslant } \right)\;k\,\,\left( { \leqslant \,m} \right)} {\left( \begin{gathered} m \\ k \\ \end{gathered} \right)C(m - k,n,l)} = \sum\limits_{\left( {0\, \leqslant } \right)\;k\,\,\left( { \leqslant \,m} \right)} {\left( \begin{gathered} m \\ k \\ \end{gathered} \right)C(k,n,l)} \tag {2} $$

which is satisfied by (1).
This implicit relation can be inverted, using the binomial inversion theorem, giving

$$ \begin{gathered} C(m,n,l) = \sum\limits_{\left( {0\, \leqslant } \right)\;k\,\,\left( { \leqslant \,m} \right)} {\left( { - 1} \right)^{m - k} \left( \begin{gathered} m \\ k \\ \end{gathered} \right)\left( \begin{gathered} n\,k \\ l \\ \end{gathered} \right)} = \sum\limits_{\left( {0\, \leqslant } \right)\;k\,\,\left( { \leqslant \,m} \right)} {\left( { - 1} \right)^k \left( \begin{gathered} m \\ m - k \\ \end{gathered} \right)\left( \begin{gathered} n\,\left( {m - k} \right) \\ l \\ \end{gathered} \right)} = \hfill \\ = \sum\limits_{\left( {0\, \leqslant } \right)\;k\,\,\left( { \leqslant \,m} \right)} {\left( { - 1} \right)^k \left( \begin{gathered} m \\ k \\ \end{gathered} \right)\left( \begin{gathered} n\,\left( {m - k} \right) \\ l \\ \end{gathered} \right)} \hfill \\ \end{gathered} \tag {3} $$

which provides a further confirmation of the validity of (1).

Referring to the scheme, we can infer an additional recurrence.
Consider to add an empty box aside the others, and move one ball at time from the original block to the new box. Each configuration obtained is a partition of that corresponding to $m+1,n,l$, and the No. of rows is accounted by $C(m,n,l-k) C(1,n,k)$.
The process goes on till the new box is full ($k=n$) or the original block is empty ($k=l$). But we can waive these thresholds, as they are implicit in the product of the two $C$ factors, defined as in 1).

So we can write

$$ C(m + 1,n,l)\quad \left| {\;0 \leqslant \text{integer}\;m,n,l} \right.\quad = \sum\limits_{\left( {0\, \leqslant } \right)\;k\,\,\left( { \leqslant \,l} \right)} {C(m,n,l - k)\;C(1,n,k)} \tag {4} $$

which is a convolutory recurrence and which is satisfied by formula 1).
In fact, considering the layout of the boxes, it is clear that a more general recurrence holds: $$ C(m + q,n,l)\quad \left| {\;0 \leqslant \text{integer}\;m,n,l,q} \right.\quad = \sum\limits_{\left( {0\, \leqslant } \right)\;k\,\,\left( { \leqslant \,l} \right)} {C(m,n,l - k)\;C(q,n,k)} $$

To solve this recurrence autonomously, without recurring to the conjectured formula 1), it remains to fix the initial conditions.
In the case of a single box, by definition $C(1,n,l)$ clearly is:

$$ C(1,n,l)\quad = \quad \begin{array}{*{20}c} {n\backslash l} &| & 0 & {1 \leqslant l} \\ \hline 0 &| & 0 & 0 \\ {1 \leqslant n} &| & 0 & {\left( \begin{gathered} n \\ l \\ \end{gathered} \right)} \\ \end{array} \quad \quad = \quad \left( \begin{gathered} n \\ l \\ \end{gathered} \right) - \left( \begin{gathered} 0 \\ l \\ \end{gathered} \right) $$

then it is easy to check that the recurrence 4) gives as solution formula 1).

Note that in the case of having no box, i.e. $m=0$, formula 1) returns

$$ C(0,n,l)\quad = \quad \begin{array}{*{20}c} {n\backslash l} &| & 0 & {1 \leqslant l} \\ \hline 0 &| & 1 & 0 \\ {1 \leqslant n} &| & 1 & 0 \\ \end{array} \quad \quad = \quad \left( \begin{gathered} 0 \\ l \\ \end{gathered} \right) $$

i.e. that the "empty" box is filled with $0$ balls (which is a usual concept in combinatorics) .
So $C(m,n,l)$ counts the arrangements with no empty box, and in case of keeping the definition of with at least 1 ball per box then $m$ should be limited to be greater than $0$, or amended accordingly.

Answer 2

I don't know if this should be just a comment or an answer but anyways.

The first method leads to over counting because you count certain cases multiple times.

Example - You have 3 balls $B_1,B_2$ and $B_3$.

Say, you put $B_1$ in $C_1$ and $B_2$ in $C_2$ initially i.e. in the first step and then assigned $B_3$ to $C_2$ and in the other case $B_3$ in $C_2$ initially and then assigned $B_2$ to $C_2$.

The first method will count this twice whereas it is just one case.

The idea for the second method seems absolutely fine.