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I am trying to figure out the following combinatorics problem:

We have an even number $N$ of identical boxes, each of them labelled by $j=1,...,N$, where for each $j$ we can put only a single ball. There are three distinct types of balls, which are labelled with a value $n_{j}=0,1,2$ for a ball at the box $j$. Now the question is, how many possible combinations can we make so that the sum of all ball values in each of the boxes gives $\sum_{j=1}^{N}n_{j}=N/2$ (remember $N$ is even)?

If we make the same problem with $n_{j}=0,1$, i.e. only two different balls, the answer is the known: \begin{eqnarray} C_{N,N/2} = \left(\begin{matrix} N\\ N/2 \end{matrix}\right)=\frac{N!}{\frac{N}{2}!\frac{N}{2}!} \end{eqnarray} In that case, the problem is equivalent to asking, for a binary string of zeros and ones, how many combinations can we make such that the total sum of the element of the string is equal to $N/2$. But for the "three-state" string, I don't know how this generalizes.

Zarathustra
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  • There likely isn't a "formula" for this answer, as opposed to the zero-one case. In case no formula exists, what are you looking for in an answer? – Mike Earnest May 11 '22 at 16:14
  • I would like to confirm a numerical result for it where for N=12 I obtain a total of 8074 combinations. So I wanted to compare this result by making use of the formula (if it exists). – Zarathustra May 11 '22 at 16:19
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    This Mathematica code confirms your $N=12$ result. You can edit the value of $N$ in the code, works for up to $N=20,000$. Also, your question is a special case of this one: https://math.stackexchange.com/q/553960 – Mike Earnest May 11 '22 at 16:31
  • Thanks a lot for the code and the link, it is actually nice to get to see how this operates on a general level. – Zarathustra May 11 '22 at 16:35
  • No problem. BTW, contrary to my first comment, Marc van Leeuwen gives a formula involving a summation of products of binomial coefficients in his answer. – Mike Earnest May 11 '22 at 16:37

1 Answers1

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If you have 3 symbols, the corresponding multinomial coefficient is $$ \binom{N}{N/3,N/3,N/3} $$ and the interpretation is that this is the collection of strings with equal numbers of each symbol, assuming $N$ is a multiple of 3.

You cannot capture the equidistribution with one real valued equation only (as in the sum of ball labels is $N/2$).

However if the number of symbols is prime, say $p$, you can take the labels to be the powers of a primitive complex root of unity $\omega,\omega^2,\ldots,\omega^p=1,$ and you can let $$ S(A_1,\ldots,A_p):=\sum_{k=1}^p A_k \omega^{k}=0 $$ where $A_k$ is the number of balls labelled with $\omega^k.$ Then $S(A_1,\ldots,A_p)=0$ if and only if each label occurs equally often.

kodlu
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