The number of symmetric matrices of order 5 with each element either 0 or 1

Question

Question is to find The number of symmetric matrices of order 5 with each element either 0 or 1 .

What i am trying is

If i take matrix of order 2

$$A=\left[\matrix{ A & B \\ B & C \ }\right]$$

Here i am having $2.2.2$ cases in total .Assuming A and C are different .Is thi correct way to generalise to higher order ? Thanks

Hint: You just have to see the possible numbers for all the matrix elements in the upper right corner (including diagonal)... You have a choice between $0$ and $1$ for $5+4+3+2+1$ elements... — Martigan, Nov 19 '14 at 12:22
@Martigan So for 5 order i will have $2^{15}$ cases ,Right ? — godonichia, Nov 19 '14 at 12:25
In fact for a matrix of dimension $k$, you will have $2^{k(k+1)/2}$ matrices. — Martigan, Nov 19 '14 at 12:29
Yes, since you have no choice for the lower left corner AND for the diagonal... — Martigan, Nov 19 '14 at 12:35

score 2 · Answer 1 · edited Apr 13 '17 at 12:20

Yes your argument is a good start. You may count $n$ the amount of possibly different elements of an upper triangular matrix and then you get $2^n$ binary symmetric matrices. In the case of matrices $3\times 3$ we have $$\begin{pmatrix} A_1 & A_2 & A_3 \\ & A_4 & A_5 \\ & & A_6 \end{pmatrix}$$ where $A_1,A_2,A_3,A_4,A_5,A_6 \in \{0,1\}$ so you'll have $2^{6}$ possible matrices (note that the other elements of the matrix are fixed by symmetry). Now, for finding $n$ you may have a look here.

The number of symmetric matrices of order 5 with each element either 0 or 1

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