Boys and Girls Alternating

Read the row from left to right. No two children of the same sex may be adjacent, so the sexes must alternate. With five of each in ten seats, the row is therefore either $BGBGBGBGBG$ or $GBGBGBGBGB$. That gives exactly $2$ patterns. In the first, the boys occupy seats $1,3,5,7,9$; in the second, they occupy seats $2,4,6,8,10$.

Fix a pattern. The five boys can be placed in their five seats in $5!$ ways. Independently, the five girls can be placed in theirs in $5!$ ways. Hence

Generalisation. With $n$ boys and $n$ girls the count is $2\,(n!)^2$. With $n+1$ boys and $n$ girls, the boys must take both ends, so there is only one pattern and the count is $(n+1)!\,n!$.