[Theory-reading] Exact Combinatorial Value

[Tyson Williams <tdw@xxxxxxxxxxx>]

As Siddharth said, the exact number of (i,j,k) triples such that i+j+k=q-1 does not matter, only that it is of order q^2.

If anyone does cares about the exact value, there are (q+1 choose 2) triples. We can partition the q-1 items into three groups by inserting two markers into a list of q-1 items. There are q-1 identical items and 2 identical markers, so the number of unique permutations is (q+1)!/[(q-1)! * 2!] = q(q+1)/2 = (q+1 choose 2).

-- 
Tyson Williams
www.cs.wisc.edu/~tdw