You have \(5\) free entries along the diagonal. Since \(a_{ij}=a_{ji}\), 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 \(4+3+2+1=10\) entries.
So the number of possibly symmetric binary matrices is \(2^5\times2^{10}=2^{15}\).
