Vietnam Mathematical Olympiad

Recall that the number of natural solutions of the equation
$$\sum_{i=1}^{n} x_i = m$$
is $\binom{m+n-1}{n-1}$. We will use this fact to calculate the number of the arrangements of boys and girls satisfying the given conditions.

Let $x_i$ be the number of boys standing between $G_{i-1}$ and $G_i$ for $i \ge 2$; $x_1$ be the number of boys standing on the left of $G_1$, and $x_6$ be the number of the boys standing on the right of $G_5$. Then we have $3 \le x_2$, $1 \le x_4 \le 4$, and
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 12. \qquad (9)$$
Replacing $y_i = x_i$ for $i \ne 2$ and $y_2 = x_2 - 3$, we have
$$y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 9,$$
where $y_i \ge 0$ and $1 \le y_4 \le 4$. Putting $y_4 = 1, 2, 3, 4$ into (9), we conclude that the number of solutions $(x_1, x_2, x_3, x_4, x_5, x_6)$ satisfying (9) is
$$\sum_{y_4=1}^{4} \binom{9-y_4+5-1}{5-1} = \binom{13}{5} - \binom{9}{5} = 1161.$$
Since we can permute the boys in the row, the total number of arrangements of boys and girls satisfying the given conditions is $12! \times 1161$.