Find the number of ways in which $n$ boys and $n$ girls ($n>1$) can be seated in a row of seats if the boys and the girls are to have alternate seats and one particular boy and one particular girl must not sit next to each other. 

P.S.: Official A-level Pure Maths (1969) Answer is $2(2(n!)-1-2(n-1))$ but my answer is $2((n!)^2-1-2(n-1))$, what's wrong here?

After my deep thought, the official answer makes no sense to me. However, my answer is $2((n!)^2-(n-1)-2(n-1)(n-1)^2)$. Simply listing all possible permutation, which has $(n!)^2$ ways, and cancels those where the boy and the girl sitting next to each other, which has $(n-1)+2(n-1)^3$ ways.