12. Three girls $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$, and nine boys are to be lined up in a row. Let $n$ be the number of ways this can be done if $\mathrm{B}$ must lie between $\mathrm{A}$ and $\mathrm{C}$, and $\mathrm{A}, \mathrm{B}$ must be separated by exactly 4 boys. Determine $\lfloor n / 7!\rfloor$.
12. Answer: 3024.
Let $\Phi$ be the set of arrangements of these girls and boys under the condition that $A, B$ must be separated by exactly 4 boys. Form a block $P$ with $A, B$ at the two ends and exactly 4 boys between them. The number of ways to form such a block $P$ is
$$
2 \times\binom{ 9}{4} \times 4!\text {. }
$$

Then we have
$$
|\Phi|=2 \times\binom{ 9}{4} \times 4!\times 7!,
$$
as we can consider such a block $P$ as one item, and there are still 5 boys and one girl (i.e., $C$ ).

Note that in any arrangement in $\Phi, C$ is outside the block between $A$ and $B$. In exactly half of the arrangements of $\Phi, B$ is between $A$ and $C$. Hence
$$
n=|\Phi| / 2=\binom{9}{4} \times 4!\times 7!.
$$

Hence the answer is $\left({ }_{4}^{9}\right) \times 4!=3024$.