the number of different nxn symmetric matrices with 4 elements where a!=b!=c!=d

Question

anonymous · Answer

Hello

anonymous · Answer

Would it be something like:

  $4^{(n^2 - n) / 2 + n} = 4^{(n^2+n)/2}$

?

The idea is that there is a total of
(i) $n^2$ elements in a $n$ x $n$ matrix; and
(ii) $n$ elements along the diagonal

So subtract the diagonal, half the result, and then add back the diagonal should yield the total number of "cells".  Since each cell has 4 possible values {a,b,c,d}, raising 4 to the number of cells is therefore the answer.  Thoughts ?

anonymous · Answer

@hanson.char: I understand the approach you're taking, but I don't think the answer is correct. If you are allowing each cell to take on any of the four possible values a, b, c, d, then the possibility exists that (for example) you produce a matrix in which all cells have the value a. But such a matrix does not satisfy the conditions of the original problem, so you shouldn't be counting it in your result.

anonymous · Answer

I probably don't fully understand the conditions of the original problem.  Is the matrix supposed to have exactly four cells with the unique values {a,b,c,d} and with all other cells set to zeros ?

anonymous · Answer

I read the problem as saying that you had an n by n symmetric matrix in which all four elements a, b, c, and d, appeared as values in the n^2 entries, with no other values present at all.

anonymous · Answer

An $n$ x $n$ matrix always contains $n^2$ elements, even if some of them are zeros.  In that sense, I don't understand what it means by "no other values present at all".