5. There are 7 boys and 3 girls standing in a row for a photo, and any two girls want to be adjacent. Then the number of possible arrangements is ().
(A) $\frac{8!}{5!}$
(B) $\frac{7!6!}{4!}$
(C) $\frac{10!3!}{7!}$
(D) $\frac{10!7!}{3!}$
5. (A).

If we think about it using the insertion method, we can easily get the answer as $C_{8}^{3} \cdot P_{7}^{7} \cdot P_{3}^{3}$ $=\frac{8!}{5!}$.

First, let the 7 boys stand in a row. At this point, the number of arrangements is $P_{7}^{7}$. Then, from the 8 possible positions (the spaces between the boys or the ends), choose 3 positions, which can be done in $C_{8}^{3}$ ways; finally, let the 3 girls occupy these 3 chosen positions, which can be done in $P_{3}^{3}$ ways; by the multiplication principle, we get the answer.