1

According to this paper, MacMahon's formula for finding $k$-combinations in finite multisets is given as follows. Let $A = \{m_1 \cdot a_1, m_2 \cdot a_2, ..., m_n \cdot a_n \}$ be a multiset, in which the multiplicities $m_i, \ i=1,...,n$ are finite. Let $C(k; m_1, m_2, ..., m_n)$ denote the number of possible $k$-combinations in $A$. Then

$$ C(k; m_1, m_2, ..., m_n) = \sum_{p=0}^n (-1)^p \sum_{1\leq i_1 \leq i_2 \leq ... \leq i_p \leq n} {n+k-m_{i_1} - m_{i_2} - ... - m_{i_p} - p - 1 \choose n+k-m_{i_1} - m_{i_2} - ... - m_{i_p} - p}. \tag{1}$$

I'm having a hard time trying to make out what the last bit means, as the upper number in the binomial coefficient is always smaller than the lower one. I know generalizations for the binomial coefficient exist, but even with them I'm not getting correct results.

The paper goes on and states that an equivalent formula would be

$$ C(k; m_1, m_2, ..., m_n) = \sum_{p=0}^n (-1)^p \sum_{1\leq i_1 \leq i_2 \leq ... \leq i_p \leq n} {n+k-m_{i_1} - m_{i_2} - ... - m_{i_p} - p - 1 \choose n-1}. \tag{2}$$

where the summation is taken for terms in which $n+k-m_{i_1} - m_{i_2} - ... - m_{i_p} - p>0$. But that doesn't make it any clearer, as the problem with the binomial coefficient still persists.

I couldn't find MacMahon's original publication, but on searching online I've found another paper on the topic, in which not only the formula (2) seems to be slightly diferent (with the indices $i_j$ strictly smaller than the next one) but which also states that the formulas (1) and (2) are nowhere to be found in the reference given in the fisrt paper.

Can anyone point me in the right direction in regards to the interpretation of that binomial coefficient?

Cheers, I.

  • As far as I can tell, the equation from the first paper (Juric and Siljak) is dead wrong, while the second paper (Wieder) is correct. There are different proofs in this question: https://math.stackexchange.com/q/553960/177399 – Mike Earnest Apr 21 '21 at 22:47
  • Fantastic, thank you. How should I close this question? – impresso Apr 23 '21 at 21:02
  • You could post an answer citing the other answer. I would post an answer myself, but in order for that answer to be convincing, I would need to find a counter-example to the first formula, but I would rather not try to do that. – Mike Earnest Apr 26 '21 at 21:52

0 Answers0