<s>[INST] How many ways are there to arrange students and teachers in two lines under the given conditions? There are $24$ children and $4$ teachers. There are $2$ pairs of twins within the 24 children.  We want to arrange everyone in two lines.  In each line, there will be $12$ children, their order of appearance (or just order) does not matter, but there will be $1$ pair of twins in each line, there will be $2$ teachers in each line, on the sides of the line, and their order DOES MATTER.  How many possibilities are for arrangement are there?
What I did is pick $10$ out $20$ children, without/ rep & without/ order * pick $1$ out of $2$ couples of twins without/ rep & without/ order, pick $2$ out of $4$ teachers WITH order and without rep, and pick $2$ out of $2$ with order and without rep.  This all leads to 
$$\frac{20!}{10!10!} \cdot 2 \cdot 2 \cdot 3 \cdot 2 \cdot 2$$
The answer is $$\frac{20!}{10!10!}  \cdot 2 \cdot 2 \cdot 3 \cdot 2$$ 
Can someone explain whats wrong with my solution? thanks
 [/INST] Consider the line containing a particular teacher.
There are  $\begin{pmatrix}20\\10\\\end{pmatrix}$ ways of choosing the children in this line and, for each of these choices, $\begin{pmatrix}2\\1\\\end{pmatrix}$ ways of choosing the pair of twins and $\begin{pmatrix}3\\1\\\end{pmatrix}$ ways of choosing the fellow teacher. The teachers can then be arranged in $2\times2$ ways giving the answer $$\begin{pmatrix}20\\10\\\end{pmatrix}\begin{pmatrix}2\\1\\\end{pmatrix} \begin{pmatrix}3\\1\\\end{pmatrix}\times2\times2=\frac{20!}{10!10!}  \cdot 2 \cdot 2 \cdot 3 \cdot 2.$$
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