You have [tex]5[/tex] free entries along the diagonal. Since [tex]a_{ij}=a_{ji}[/tex], you only need to worry about how many choices are available for the triangular part either above or below the diagonal. These parts both contain [tex]4+3+2+1=10[/tex] entries.
So the number of possibly symmetric binary matrices is [tex]2^5\times2^{10}=2^{15}[/tex].
