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This appears to be an inclusion/exclusion problem. My first step was to find the total permutations with no restrictions, using $\frac{8!}{2!2!2!2!} = 2520$.

What would be the permutation formulas for all adjacent A's, B's, C's, and D's? Furthermore, how do I know what is added and subtracted?

Ludwigthestud
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2 Answers2

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To count the cases where, say $AA$ and $CC$ (and possibly more) are adjacent, count the permutations of $\bar A,B,B,\bar C,D,D$ and imagine barred letters stand for double letters.

Hagen von Eitzen
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  • Ok, so in essence I treat each pair as a single space, and then for the remaining 6 spaces I have 2 choices. So for the choices of A, it is found with $\frac{7!}{2!2!2!}$. Do I do the same equation for the remaining three letters? – Ludwigthestud Sep 24 '18 at 17:38
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First lets find the total number of permutations without the given condition.

The total number of permutations are $N=\dfrac{8!}{2!2!2!2!}=2250$

Now, by applying the inclusion exclusion principle, a permutation has a property $\alpha$ in case $A's$ are adjacent, and $\beta$ for $B's$ and $\gamma$ for $C's$

$$N(\alpha)=\dfrac{7!}{2!2!2!}=630$$$$N(\alpha,\beta)=\dfrac{6!}{2!2!}=180$$$$N(\alpha,\beta,\gamma)=60$$$$N(\alpha,\beta,\gamma,\delta)=24$$ Therefore,$$N-4N(\alpha)+6N(\alpha,\beta)-4N(\alpha,\beta,\gamma)+N(\alpha,\beta,\gamma,\delta)=864$$

Key Flex
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  • Ok, here's what confuses me. Why is it $N - 4(630) + 6(180) -4(60) + 24$? In other words, why is each result be multiplied? – Ludwigthestud Sep 24 '18 at 18:09