Re: [Theory-reading] Exact Combinatorial Value

[Siddharth Barman <sid@xxxxxxxxxxx>]

Attached pdf has a more comprehensive coverage of the general problem.
-Siddharth

Tyson Williams wrote:
> 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 <http://www.cs.wisc.edu/%7Etdw>
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Attachment: lecture15.pdf
Description: Adobe PDF document