Find number of ways to seat $n$ boys and $n$ girls in a row so that every boy has at least one girl sitting beside him. My attempt: I am getting $2^n(n!)^2$ .
First I paired $n$ boys and $n$ girls in $n!$ ways then these pairs can be arranged in $n!$ ways and in each of these pairs boy and girl can arrange themselves in $2!$ ways.

With a perspective of keeping  it simple and $z$ representing boys and
$w$ representing girls we get the generating function
$$(1+z)
\left( \sum_{q\ge 0} \left(\frac{w}{1-w} (z+z^2)\right)^q \right)
\frac{w}{1-w} (1+z).$$
This is
$$(1+z)^2 \frac{w}{1-w} \frac{1}{1-w(z+z^2)/(1-w)}
= w (1+z)^2 \frac{1}{1-w-w(z+z^2)}.$$
Extracting coefficients from this we get
$$[w^n] w (1+z)^2 \frac{1}{1-w-w(z+z^2)}
= (1+z)^2 [w^{n-1}] \frac{1}{1-w(1+z+z^2)}
\\ = (1+z)^2 (1+z+z^2)^{n-1}
= (1+z+z^2)^{n} + z (1+z+z^2)^{n-1}.$$
This yields as the answer the two trinomial coefficients
$$[z^n] (1+z+z^2)^{n} + [z^n] z (1+z+z^2)^{n-1}
\\ = [z^n] (1+z+z^2)^{n} + [z^{n-1}] (1+z+z^2)^{n-1}.$$
Extracting coefficents we get for the first term
$$[z^n] \sum_{q=0}^n {n\choose q} z^q (1+z)^q
= \sum_{q=0}^n {n\choose q} {q\choose n-q}$$
for a final answer of
$$\sum_{q=0}^n {n\choose q} {q\choose n-q}
+ \sum_{q=0}^{n-1} {n-1\choose q} {q\choose n-1-q}.$$
Given that we have an excellent answer the above can perhaps provide a
slightly   different  perspective,   thereby  facilitating   a  better
understanding of the computation.
