Five boys and five girls form a line with the boys and girls alternating. Find the ways of making the line. The way I solved this problem was: First, let's place the boys standing in line.
_ B _ B _ B _ B _ B _
So there are six gap places left where the girls can stand in $_6P_5$ ways and the boys can be arranged in $5!$ ways.
So according to me the answer should be $5!$ ($_6P_5$). But the answer was $2(5!)^2$.
I don't understand where did I go wrong.
The boys aren't identical first of all, so you need to take that into account. Secondly you can't choose those six places for the girls to line up; some of those choices will leave two boys next to each other. Thirdly, $(_{6}P_{5})$ is not equal to 5!. I'm not sure $(_{6}P_{5})$ would be what you want in this situation anyway; it would mean you have 6 objects placed into 5 places without repetition.
(I think I misunderstood some things in the original version of the question. But the main problem remains that the five girls cannot be arbitrarily placed into the six positions you have for them without sometimes leaving boys next to each other.)