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Let we have 4 boys and 4 girls and we need to arrange them, with no two boys and no two girls are next to each other? Let place first the boys

b b b b

the 4 boys can arrange themselves by 4! ways. Now we have 5 positions between boys but we have 4 girls. If we arrange them like $5^{P}4 = 5!$ then the total arrangement will be = $5! \times 4!$

But there is a problem,

If we think, g b g b g b (*) b

In the above style we have a boy and boy repetition. How to solve this problem?

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    If no two boys can be adjacent, and no two girls can be adjacent, the arrangement have to alternate BGBGBGBG, or GBGBGBGB. There are $4!^2$ ways to create each configuration. – ant11 Jul 16 '14 at 17:20
  • We are having the two different patterns so , will not we have to multiply 2 ? = $2\times 4! \times 4! $ ? –  Jul 16 '14 at 17:23
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    Yes, the final answer should be $2\times 4!^2=1152$ – ant11 Jul 16 '14 at 17:25
  • Your calculation gives the answer to the question in the title, which is not the same as the question in the body. – André Nicolas Jul 16 '14 at 17:48
  • Similar to http://math.stackexchange.com/questions/733427/there-are-5-boys-and-5-girls-find-the-ways-in-which-boy-and-girl-can-sit-al?rq=1 – Dietrich Burde Jul 16 '14 at 18:16

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