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I understand why one solution to this problem works, but I'm confused as to what is wrong with my solution. Please advise!

Question: There are $11$ players on a football team who are asked to line up in one straight line for a team photo. Three of the team members named Adam, Brad and Chris refuse to stand next to each other. Find the number of different orders the players can be positioned for this photo.

My thought process is that I just need to find the total ways and then subtract the ways where any 2 of the players stand together. Ie. any variation where $AB$, $BA$, $AC$, $CA$, $BC$, $CB$ are together. My assumption is that it doesn't matter whether the third is also next to the others, once two are together that variation needs to be subtracted regardless. So I found the number of ways any two could be together ($6\cdot 10!$)and subtracted from $11!$. But I didn't get the right answer. I understand the other way to do it but I'm curious as to what incorrect assumption I have made in this method.

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  • $\begingroup$ See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula. $\endgroup$ Commented Jul 10 at 11:41

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The problem is that you have counted situations where all three are next to each other twice each in the $6\times 10!$. For example, xxACBxxxxxx has been counted once as xx(AC)Bxxxxxx and once as xxA(CB)xxxxxx. To correct for this, you need to add all of these orderings (of which, for the same reason, there are $3!\times 9!$) back on.

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  • $\begingroup$ Thank you so much! $\endgroup$
    – Jess
    Commented Jul 10 at 12:30

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