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Enumerative combinatorics, graph theory, order theory, posets, matroids, designs and other discrete structures. It also includes algebraic, analytic and probabilistic combinatorics.

17 votes
2 answers
639 views

Geometric/combinatorial depiction of algebraic identity?

I'm looking for a geometric or combinatorial depiction of the algebraic identity $$ xyz = \frac{1}{24} \Big\{(x+y+z)^3 - (x-y+z)^3 - (x+y-z)^3 + (x-y-z)^3 \Big\}. \label{*}\tag{$*$} $$ Here is the …
Zach Teitler's user avatar
  • 6,237
13 votes
Accepted

Sum of multinomals = sum of binomials: why?

For convenience set $m=n-2k$. Then \begin{equation} \begin{split} \binom{n-2k+j}{j,k-2j,n-3k+2j} &= \binom{m+j}{j,k-2j,m-k+2j} \\ &= \binom{m+j}{m} \binom{m}{k-2j} \\ &= [t^j](1-t)^{-(m+1)} …
Zach Teitler's user avatar
  • 6,237
11 votes
0 answers
252 views

Poset of nonvanishing minors of a matrix

This question was posed on MSE here three days ago, but hasn't gotten any answers or suggestions. I hope it's okay to ask it on MO, but if I should wait a little longer, please just let me know. Say …
Zach Teitler's user avatar
  • 6,237
7 votes

Tensor rank of anti-symmetric tensor

The tensor ranks of determinants and permanents are currently not known. In the $3 \times 3$ case it is known: the $3 \times 3$ determinant has tensor rank $5$ and the $3 \times 3$ permanent has tenso …
Zach Teitler's user avatar
  • 6,237
5 votes
Accepted

Number of sets containing some given sets of fixed cardinality

Let $I = \{(T_j,T') : 1 \leq j \leq L, T_j \subseteq T', |T'|=m\}$. Projection on the first factor is $\binom{n-k}{m-k}$-to-$1$, so $|I|=\binom{n-k}{m-k}L$. Projection on the second factor is at most …
Zach Teitler's user avatar
  • 6,237
5 votes

Reference request: A multidimensional generalization of the fundamental theorem of calculus

The $p=2$ dimensional case is an exercise in Rogawski's calculus textbook. It is exercise 47 on page 885, section 15.1 (Integration in Several Variables) in the 2008 Early Transcendentals edition.
Zach Teitler's user avatar
  • 6,237
5 votes
Accepted

Combinatorial proof of identity

Santa Claus has $N+1$ reindeer whose noses are of varying redness. Every year, Santa needs $n+1$ reindeer to pull his sleigh. The reddest-nosed reindeer always leads the sleigh. The way Santa chooses …
Zach Teitler's user avatar
  • 6,237
3 votes
Accepted

Covering all except one of the purple intersection points of $n$ red and $m$ blue lines effi...

The very nice paper "Cayley-Bacharach theorems and conjectures" by David Eisenbud, Mark Green, and Joe Harris (http://www.ams.org/journals/bull/1996-33-03/S0273-0979-96-00666-0/home.html) gives an int …
Zach Teitler's user avatar
  • 6,237
2 votes
Accepted

Partitions of finite sets and their behavior under permutations of the set

(Not an answer, but too long for a comment.) What if $X = \{1,2,3,4,5,6\}$, $A=\{1,2,3\}$, and $B=\{4,5,6\}$; and $\sigma_1=(3,6)$, $\sigma_2=(2,6)\sigma_1^{-1} = (2,6,3)$. Anyway $\sigma_2 \sigma_1 = …
Zach Teitler's user avatar
  • 6,237
2 votes

Sum of products of binomials

Let $G$ be the (infinite) graph with vertex set $\mathbb{Z}^2$, and the following edges. When $x+y < 0$, the vertex $(x,y)$ has outgoing edges to $(x+1,y)$ and to $(x,y+1)$. When $x+y \geq 0$, the ver …
Zach Teitler's user avatar
  • 6,237
2 votes
Accepted

Given the index of two permutations, Is there a direct way to compute the index of their com...

As noted in the comments, there are problems with adding leading zeros to an index (indices $10$, $010$, $0010$ correspond to different permutations $21$, $132$, $1243$) or conversely, adding fixed po …
Zach Teitler's user avatar
  • 6,237
1 vote

Examples of ubiquitous objects that are hard to find?

Explicit tensors or polynomials of general rank. The Waring rank of a general polynomial (meaning, general coefficients), in a given number of variables and of a given degree, has been known since 1 …
0 votes

Closed form of $\sum_{i=k}^\infty i h {i \choose {k-1}} h^{k-1} (1-h)^{i - (k-1)}$?

$$ \begin{split} \sum_{i=k}^{\infty} i h \binom{i}{k-1} h^{k-1} (1-h)^{i-k+1} &= h^k (1-h)^{2-k} \sum_{i=k}^{\infty} i \binom{i}{k-1} (1-h)^{i-1} \\ &= h^k (1-h)^{2-k} f(1-h), \end{split} $$ …
Zach Teitler's user avatar
  • 6,237