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There are $5$ boys and $5$ girls. Find the ways in which boy and girl can sit alternately.
I think it is $$5!×\binom65\times5!$$ I used this method: First let the boys sit in 5 places: BBBBB and they have $5!$ ways to be arranged. And we have 6 places left for girls as _B_B_B_B_B_
Where am I wrong?
It isn't 6 choose 5. In order to actually sit alternatively, there are only two such ways; either GB GB GB GB GB or BG BG BG BG BG. Thus, it is $2(5!)^2$.
