If you are interested in approximations to your ratios, you may find the accepted answer (and some comments of mine) to this MathOverflow post useful: Sum of 'the first k' binomial coefficients for fixed …

Aug 17, 2015 · Zaslavsky's formula is a very important formula in enumerative combinatorics, as well as geometric combinatorics, and the basis for important developments in topological combinatorics.

In addition to the OP's 2011 paper with Ž. Jurić: A New Formula for the Number of Combinations of Permutations of Multisets Applied Mathematical Sciences, Vol. 5, 2011, no. 18, 875-881 there is a …

Feb 17, 2025 · How to prove that $$\\sum_{b=0}^{[\\frac{n}{2}]}(-1)^{n-b-1}(n-b-1)!\\binom{n}{n-2b}\\frac{\\binom{2b}{2}\\binom{2b-2}{2}\\dots \\binom{2}{2}}{b!}2^{n-b}$$ equals ...

One can take this a step further. In addition to combining pairs of terms of the original sum N choose i to get a sum of terms of the form N+1 choose 2j+c, where c is always 0 or always 1, one can now take …

First component (for i = 0) is just a regular combination with repetition, then I subtract all that have at least 1 box overfilled, then I need to add those that have at least 2 box overfilled (since earlier I …

I am trying a variation of non-repetitive combination scenario. I can use the formula n!/r!x (n-r)! to find non-repetitive combinations of size "r" from "n" numbers.

The Wikipedia page List of probabilistic proofs of non-probabilistic theorems has a reference to the paper: Blyth, Colin R.; Pathak, Pramod K. A Note on Easy Proofs of Stirling's Theorem. Amer. Math. …

Apr 2, 2010 · The Weyl character formula tells us how to go from the former to the latter. My question is: is there much known about the matrix of going from the latter to the former? I've gone through a few …

Oct 26, 2015 · Rather than expressing this as a linear combination with of $1,n,n^2,\cdots,n^ {2q}$ with rational coefficients, it is better to express it as a linear combination of $\binom {n} {0},\binom {n} …