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Suppose $5$ boys and $4$ girls are to be arranged in a queue such that between any two boys there is at least one girl. Find the number of such arrangements possible.

What i think is $5$ boys can be arranged in $5!$ ways at $5$ places in the queue.Now,since it is given that at least $1$ girl must be present between any two boys,then the only case possible is : B G B G B G B G B as this is only condition when there would be $1$ girl present between any two boys.But the girls can make arrangements in these $4$ places in $4!$ ways. So, the total number of possible arrangements are $5!4!$...am i thinking it right??? could anyone tell if i am wrong somewhere in my thinking process...and if i am right please do let me know....any help is much appreciated...

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  • $\begingroup$ Seems right to me. $\endgroup$ – Ehsan M. Kermani May 10 '13 at 6:14
  • $\begingroup$ thanks for your feedback...i just wanted to know if you are sure about it...thanks :) $\endgroup$ – under root May 10 '13 at 6:20
  • $\begingroup$ yes you are going right.answer will 5!*4!. $\endgroup$ – iostream007 May 10 '13 at 6:31
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The boys and girls must be placed in an arrangement like the following. $$\require{wasysym}\begin{array}{ccccccccc} \color{darkblue}{b_1} & \color{lightpink}{g_1} & \color{darkblue}{b_2} & \color{lightpink}{g_2} & \color{darkblue}{b_3} & \color{lightpink}{g_3} & \color{darkblue}{b_4} & \color{lightpink}{g_4} & \color{darkblue}{b_5} \end{array}$$ We can rearrange the $\color{darkblue}{\text{boys}}$ and $\color{lightpink}{\text{girls}}$ independently, and there is no restriction upon arrangement within those sets. Therefore the number of admissible arrangements is $$\color{darkblue}{5!}\times\color{lightpink}{4!}=2880.$$

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