### for catherine j.

a recent post in KTM asked why
${6 \choose 2} = {6 \choose 4}$.
i know that one.

the “graph” in the upper right is K_4…
the “complete graph on 4 vertices”…
and has *6* (so-called) edges.

its “complement” (upper left) has *none*
of the edges. obviously there’s one
(so-called) *complete* graph having
all six possible edges and one “empty”
graph having *none* of the edges.

move down to the next couple lines:
the graphs having exactly *one* edge
match up in one-to-one fashion
with the graphs having *five* edges
(because “including five (edges)”, in this context,
is the same as “leaving out *one* (edge) out”.

so on for two-edge graphs.
each is the “complement”
(via the inclusion-exclusion principle
as hinted at a moment ago)
of a *four*-edge graph.

posting; gotta go to class.

1. You are completely backwards. The black “curves” are empty spaces where edges should go. Once they are filled in, once the edge is there, you can no longer see it.

So the second diagram down on the left has 4 edges, and the 2nd down on the right has 2.

(ok, that seem wiseguy-ish? So why does my logic text color IN Venn diagrams to indicate empty regions?)

Jonathan

• ## (Partial) Contents Page

Vlorbik On Math Ed ('07—'09)
(a good place to start!)