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.
 A: Begin with seating $n$ indistinguishable girls leaving ample space between and around  them, like so:
$$\ \underline{\ \  }\ G\ \underline{\ \ }\ G\ \underline{\ \ }\ G\ \underline{\ \ }\ \ldots\ \underline{\ \ }\ G\ \underline{\ \ }\ G\ \underline{\ \ }\quad .$$
An admissible seating pattern can then be constructed as follows:


*

*Choose an $r$ with $0\leq2r\leq n$.

*Choose $r$ of the $n-1$ inner spaces between the $G$s, and write $B^2$ there. This can be done in ${n-1\choose r}$ ways.

*Choose $n-2r$ of the remaining $n+1-r$ spaces, and write a single $B$ there. This can be done in ${n+1-r\choose n-2r}$ ways.
The total number $P(n)$ of admissible seating patterns therefore comes to
$$P(n)=\sum_{r=0}^{\lfloor n/2\rfloor}{n-1\choose r}\>{n+1-r\choose n-2r}\qquad(n\geq1)\ ,$$
producing the sequence $(2,4,10,26,70,192,\ldots)$, as in Markus Scheuer's answer. Making girls as well as boys distinguishable then gives the number
$$N=(n!)^2 P(n)$$
of personalized seatings.
A: 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.
A: Start with two lines, one of boys and one of girls, and imagine constructing the arrangement by picking the first person in one of the lines. You get a factor of 2 every time the choice can be from either line.
This construction is useful because no seating arrangement is double counted, which can be seen by noting that you can retrieve the order the boys (or girls) were in in their monogender lines by skipping the people of another gender. 
