Q: How many ways are there to distribute $12$ red jelly beans to four children, a pair of identical female twins and a pair identical male twins? 
Question: How many ways are there to distribute $12$ red jelly beans to four children, a pair of identical female twins and a pair identical male twins?

I attempted to solve this problem. Here is my solution:
Distribute $n$ jelly beans to female twins and $12-n$ to male twins(Maybe twins are supposed to be identical if their sex is same). If $n$ is even, then the number of ways to distribute $n$ jelly beans to female twins is $\frac{n}{2}+1$ and if $n$ is odd, the number of ways to distribute $n$ jelly beans to female twins is $\frac{n+1}{2}$. Thus the number of ways to distribute 12 jelly beans to a pair of identical female twins and a pair of identical male twins is
\begin{align}
&\sum_{n\text{ even}\le 12}\left(\frac{n}{2}+1\right)\left(\frac{12-n}{2}+1\right)+\sum_{n\text{ odd}\le 12}\left(\frac{n+1}{2}\right)\left(\frac{13-n}{2}\right)\\
&=\sum_{k=0}^6 (k+1)(7-k) + \sum_{k=1}^6 k(7-k)\\
&=7+\sum_{k=1}^6 (7+6k-k^2)+\sum_{k=1}^6 (7k-k^2)\\
&=7+\sum_{k=1}^6 (7+13k-2k^2)\\
&=7+ 7\cdot 6 + 13\cdot \frac{6\cdot 7}{2} - 2 \cdot \frac{6\cdot 7\cdot 13}{6}\\
&=7+42+273-182\\
&=140.
\end{align}
However, I want to solve it by applying Polya enumeration theorem, but I have no idea.

A: Instead of Polya enumeration theorem, I found an answer using Burnside's lemma. Define sets
$$
X=\{(f_1,f_2,m_1,m_2):f_1,f_2,m_1,m_2\text{ are integers and }f_1+f_2+m_1+m_2=12\}
$$
and
$$
G=\{e,g_1,g_2,g_3\},
$$
where $e,g_1,g_2,g_3$ are functions from $X$ to $X$ such that
\begin{align}
e(f_1,f_2,m_1,m_2)&=(f_1,f_2,m_1,m_2),\\
g_1(f_1,f_2,m_1,m_2)&=(f_2,f_1,m_1,m_2),\\
g_2(f_1,f_2,m_1,m_2)&=(f_1,f_2,m_2,m_1),\\
g_3(f_1,f_2,m_1,m_2)&=(f_2,f_1,m_2,m_1).
\end{align}
$G$ is a group under function composition, because the Cayley table of $G$ is
$$
\begin{array}{c|cccc}
\circ & e & g_1 & g_2 & g_3\\
\hline
e & e & g_1 & g_2 & g_3\\
g_1 & g_1 & e & g_3 & g_2\\
g_2 & g_2 & g_3 & e & g_1\\
g_3 & g_3 & g_2 & g_1 & e
\end{array}$$
Thus we can define a group action of $G$ on $X$ by $g\cdot x=g(x)$. By Burnside's lemma, the number of orbits is
\begin{align}
&\frac{1}{|G|}(|F_{e}|+|F_{g_1}|+|F_{g_2}|+|F_{g_3}|)\\
&= \frac{1}{4}\left(\left(\binom{4}{12}\right)+2\left(\left(\binom{2}{12}\right)+\left(\binom{2}{10}\right)+\left(\binom{2}{8}\right)+\left(\binom{2}{6}\right)+\left(\binom{2}{4}\right)+\left(\binom{2}{2}\right)+\left(\binom{2}{0}\right)\right)+\left(\binom{2}{6}\right)\right)\\
&=\frac{1}{4}\left(\binom{15}{12}+2\left(\binom{13}{12}+\binom{11}{10}+\binom{9}{8}+\binom{7}{6}+\binom{5}{4}+\binom{3}{2}+\binom{1}{0}\right)+\binom{7}{6}\right)\\
&=\frac{1}{4}(455+2(13+11+9+7+5+3+1)+7)\\
&=140,
\end{align}
where $F_g=\{x\in X:g\cdot x=x\}$ and $\left(\binom{n}{r}\right)$ the number of combinations with repetition.
