Find the number of ways to arrange 7 boys and 4 girls such that no two girls are together.
My approach:
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First fix 7 boys. This can be done in $(7-1)!=6!$ ways.
This creates 7 slots for girls. -
Now these 4 girls can be arranged in $~_7P_4 = {7\choose 3}\cdot 4!=\frac{7!}{3!}$.
So total number of ways to arrange boys and girls according to given condition is $$\frac {6!\cdot 7!}{3!}$$
Is it correct? If not…please tell me how to do it.
P.S. i'm new here. I know the moderation here is pretty strict and people won't like my not using Mathjax to typeset my question.i'm sorry. I will learn it as soon as possible.
Thank you!
Best Answer
Your working and explanation are both correct.
As an aside, to further explain why when using circular permutations for the boys we only use $6!$ rather than $7!$, due to the circular nature of the table we don't try to keep track of which way was north or south, etc... To avoid overcounting we can take a specific boy, for example the youngest one, and allow him to sit anywhere at the table first before anyone else. From that point on, we can organize our thoughts about where everyone else sits in reference to him. After having seated the youngest boy, there remain only $6$ other boys to arrange around the table after him, hence the result.