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The combinatorial identity $$ \sum_{k=0}^{n-1} \binom{n+1}{k} \binom{n+1}{k+1} \binom{n+1}{k+2} = 2 \sum_{k=0}^{n-1} \binom{n+1}{k} \binom{n}{k} \binom{n+1}{k+2} $$ has a simple combinatorial proof. T …

The theorem still holds if we ask each row and column to sum to some integer $m$; permutation matrices are replaced with zero-one matrices having the same constraints. See Watkins and Merris, Convex S …

Given $w \leq n$, consider the graph $G$ whose vertex set is $F_2^n \times \binom{[n]}{w}$, and two vertices $(v,S),(v',S')$ are connected if $v|_S = v'|_S$ or $v|_{S'} = v'|_{S'}$, where $v_S \in F_2 …

There is (probably) no nice condition for a graph to be 3-colorable, since deciding whether a graph is 3-colorable is NP-complete. Moreover, assuming the Exponential Time Hypothesis, deciding whether …

There is an extremely simple method attributed to Conway by Eric Weisstein, who describes it in a MathWorld article. All formulas below assume that row and column indices are 1-based. We have to dist …

We say that a multilinear polynomial $P(x_1,\ldots,x_n)$ in $n$ commuting variables over $\mathbb{R}$ has zero trace if $$ \frac{d}{dt} P(t,\ldots,t) = 0. $$ Equivalently, $$ \left(\sum_{i=1}^n \frac{ …

The basis appears in Murali K. Srinivasan's paper Symmetric chains, Gelfand-Tsetlin chains, and the Terwilliger algebra of the binary Hamming scheme, though perhaps not as explicitly. The author shows …

There are at least three types of results that spring to mind. One is the Ahlswede-Khachatrian theorem ("the complete intersection theorem"), which for each $n$ and $k < n/2$ will give you tight upper …

If you fix $B$ then the situation is described by a DFA (deterministic finite automaton), i.e. the set of permissible words is a regular language, and so has a rational generating function; therefore, …

A step is a movement of magnitude 1, a hop of magnitude 2. Denote by $X$ a visited place, by $O$ a place not visited, and by $Y$ the current position. A self-avoiding walk hovers around the states $E …

Suppose we have n elements, we want to generate r elements (without order), and want exactly k elements to be repeated. We can separate the result into a part containing all the repeated elements (th …

Another answer (I guess they must be equivalent): Write each original line as a difference of two 0/1 vectors. Adapt this representation to the modified lines by changing only the subtrahends. You n …