How many ways are there to seat six boys and eight girls in a row of chairs so that no two boys are seated next to each other? The way I did it was I first drew out $6$ boys and put a girl between each boy. Then I made $6$ barriers so it looked like this: 
b | g b | g b | g b | g b | g b | 
From these barriers, I arranged the $3$ remaining girls into one of the slots (from $6$ barriers and 3 girls or $\binom{9}{3}$. Then, I multiplied by the number of ways to arrange the girls and boys. My final answer was 
$$\binom{6 + 3}{3} \cdot 8! \cdot 6!$$
After looking at similar questions on Stack, I was able to solve the problems using this strategy, however the answer my textbook has tells me that I'm over counting.
 A: Method 1:  We can arrange the eight girls in a row in $8!$ ways.  This creates nine spaces in which to insert the boys, seven between successive girls and two at the ends of the row.
$$\square g \square g \square g \square g \square g \square g \square g \square g \square$$
To ensure that no two of the boys sit in consecutive seats, we choose six of these nine spaces in which to insert the boys, which we can do in $\binom{9}{6}$ ways.  The boys can be arranged in the six selected spaces in $6!$ orders, so the number of seating arrangements of eight girls and six boys in which no two of the boys sit in consecutive seats is $$8!\binom{9}{6}6!$$ as you found.
Method 2:  This is a modification of your approach.
We can arrange the six boys in a row in $6!$ ways.  This creates seven spaces in which to place the girls, five between successive boys and two at the ends of the row.
$$\square b \square b \square b \square b \square b \square b \square$$ 
If $x_k$ represents the number of girls placed in the $k$th space, then 
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 8 \tag{1}$$
The requirement that no two boys sit in adjacent seats means that $x_2, x_3, x_4, x_5, x_6 \geq 1$, while $x_1, x_7 \geq 0$.  If we let $y_k = x_k - 1$, $2 \leq k \leq 6$, then $y_k$ is a nonnegative integer.  Substituting $y_k + 1$ for $x_k$, $2 \leq k \leq 6$, in equation 1 yields
\begin{align*}
x_1 + y_1 + 1 + y_2 + 1 + y_3 + 1 + y_4 + 1 + y_5 + 1 + y_6 + 1 + x_7 & = 8\\
x_1 + y_1 + y_2 + y_3 + y_4 + y_5 + y_6 + x_7 & = 3 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers.  A particular solution corresponds to the insertion of six addition signs in a row of three ones.  The number of such solutions is 
$$\binom{6 + 3}{6} = \binom{9}{6}$$
since we must choose which three of the nine positions (three ones and six addition signs) will be filled with additions signs.  Finally, we can arrange the eight girls in the selected positions in $8!$ ways, again yielding $$6!\binom{9}{3}8!$$ 
A: Using generating functions
Let $a_{i,j}$ be the number of ways to seat $i$ boys and $j$ girls in this way (i.e. no two boys adjacent), and let $$F(x) = \sum_{x, y \ge 0} a_{i,j} \frac{x^i}{i!} \frac{y^j}{j!}.$$
Let's consider the start of the row. There are four possibilities, each followed by a row satisfiying the same condition:


*

*The row is empty (no chairs, no boys, and no girls);

*The row starts with a boy and then ends (only 1 chair);

*The row starts with a boy, followed by a girl;

*The row starts with a girl.
To count rows of the first two possibilities (the base case), the generating function is just $1 + x$.
For the third possibility (start with a boy, followed by a girl) we can multiply $F(x)$ by $xy$, since we are adding one boy and one girl. $\frac{x^i}{i!}$ multiplied by $x$ is $(i+1) \frac{x^{i+1}}{i+1}$, thus taking into account that the boy we add can be any of the $i+1$ boys, and similarly for girls.
Finally, for the fourth possibility we multiply by $y$.
This altogether gives us that
$$
F(x) = 1 + x + xyF(x) + yF(x).
$$
Good! -- it looks like we can solve for $F$.
So we have
$$
F(x) = \frac{1+x}{1 - y - xy}.
$$
We want the coefficients of $F$, so we should expand this out.
We get
\begin{align*}
F(x) &= \frac{1+x}{1 - (1+x)y} \\
&= (1+x) \sum_{j \ge 0} (1+x)^j y^j \\
&= \sum_{j \ge 0} (1+x)^{j+1} y^j \\
&= \sum_{i, j \ge 0} \binom{j+1}{i} x^i y^j \\
&= \sum_{i, j \ge 0} \frac{(j+1)!}{i! (j+1-i)!} x^i y^j \\
&= \sum_{i, j \ge 0} \frac{(j+1)! j!}{(j+1-i)!} \frac{x^i}{i!} \frac{y^j}{j!}.
\end{align*}
This gives us the coefficients of our original generating function.
We have
$$
a_{i,j} = \boldsymbol{\frac{(j+1)! j!}{(j+1-i)!} = \binom{j+1}{i} i! j!}.
$$
That's the number of ways to seat $i$ boys and $j$ girls.
And this makes sense, because we see that we can order the boys in $i!$ ways, order the girls in $j!$ ways, and then choose where to insert the boys into the spaces between the girls in $\binom{j+1}{i}$ ways.
You asked about 6 boys and 8 girls. That would be
$$
a_{6,8} = \frac{9! 8!}{(9-6)!}
= \frac{9! 8!}{3!} = 2438553600.
$$
