[Math] 4 Boys & 4 Girls are to be seated in a line find number of ways , so that Boys & Girls are in alternate seats.

combinatoricspermutations

4 Boys & 4 Girls are to be seated in a line find number of ways , so that Boys & Girls are in alternate seats.

My approach:

If boys are seated in B$1$,B$2$,B$3$,B$4$ positions than at each gap between two consecutive boys a girl can sit so, there will be C$(5,4)$ ways for girls and they can be arranged in C$(5,4)$ *4! and boys too can be arranged in 4! so total number of ways are C$(5,4)$ *4! *4!

But in textbook it's answer is 2*4!*4!

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Please help me in finding my mistake and explain it too

Please note: I am just a high school student . So,please don't close my question.

Best Answer

One of the $C(5,4)$ ways you count to seat the girls fills gaps 1,3,4,5. But then boys B1 and B2 are seated next to each other, so boys and girls don't alternate. So there are not $C(5,4)$ ways to choose the spots for girls.