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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 …

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 …

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 …

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 …

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 …

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 …

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)} …

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.

(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 = …

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 …

$$ \begin{split} \sum_{i=k}^{\infty} i h \binom{i}{k-1} h^{k-1} (1-h)^{i-k+1} &= h^k (1-h)^{2-k} \sum_{i=k}^{\infty} i \binom{i}{k-1} (1-h)^{i-1} \\ &= h^k (1-h)^{2-k} f(1-h), \end{split} $$ …

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 …

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 …