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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.
24
votes
Accepted
Combinatorics Problem: $\sum _{k=0}^{s-1} \binom{n}{k}=\sum _{k=1}^s 2^{k-1} \binom{n-k}{s-k}$
A slightly less computational method is to note that both sides of the identity count the number of subsets of $\{1,\dots,n\}$ with fewer than $s$ elements. This is obvious for the left hand side. It' …
14
votes
Accepted
Can a partition free family in $2^{[n]}$ always be enlarged to one of size $2^{n-1}$?
Yes. Suppose $\mathcal{F}$ is partition-free of smaller size. Then there is some $A$ for which $\mathcal{F}$ contains neither $A$ nor its complement. It must be possible to add either $A$ or its compl …
24
votes
Accepted
Non-isomorphic graphs with bijective graph homomorphisms in both directions between them
As vertex set, take $V=V'\cup V''$, the disjoint union of two infinite sets.
For $G$, take all edges except those joining pairs of vertices from $V''$.
For $H$, add one extra edge, between a pair of …
21
votes
Accepted
Do mutually dual finite vector spaces have the same orbit cardinalities under a linear group...
Yes. In particular, it can happen that $V$ has non-zero fixed points, but $V^*$ doesn't.
For example, let $G$ be the symmetric group of degree 3 acting in the obvious way on the set $\{e_1,e_2,e_3\}$ …
2
votes
Realising matrices as Cartan matrices
I don't know a classification, but as well as needing strictly positive entries on the main diagonal (as I pointed out in a comment), you also need to rule out matrices of the form
$$\pmatrix{A&B\\0&C …
10
votes
How to calculate$ \sum \limits_{k=0}^{m-n} {m-k-1 \choose n-1} {k+n \choose n}$?
An easy combinatorial way to see that
$$\sum_{k=0}^{m-n}\binom{m-k-1}{n-1}\binom{k+n}n=\binom{m+n}{2n}:$$
The right hand side is the number of ways to pick a subset of size $2n$ from $\{1,2,\dots,m+n …
8
votes
An operation on binary strings
I think it's less confusing if you swap the roles of 0 and 1, as then the basic operation you're using to generate the entries of $\alpha\times\beta$ is addition mod 2.
Then, if you write $\alpha_i$ …
8
votes
Accepted
On decomposition of finite Abelian groups
I don't think it's true for $G=\mathbb{F}_2^3$ and $a=b=3$.
If there were such sets $A$ and $B$, they must have exactly three elements each.
By applying a translation and a group automorphism, we ma …