# What is the number of rearrangements of the string AAABBBCCC that do not contain three consecutive letters of the same type?

Just a little combinatorial theory problem I am having trouble wrapping my head around. It has to do with rearrangement of a string of letters:

Determine the number of rearrangements of the string AAABBBCCC that do not contain three consecutive letters of the same type

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It's probably easier to determine the number that do. If all three As are in a row, there are $7$ positions for the As and $\binom63=20$ positions for the Bs and Cs, for a total of $140$. The same for all three Bs and all three Cs, makes 420. Now we've double-counted the ones where all As and all Bs are in a row – that leaves $\binom53=10$ positions for the Cs, times $2$ for whether the As or the Bs come first, and the same again for all As and all Cs and for all Bs and all Cs, so substract $60$. But now we've triple-counted and triple-uncounted the $6$ possibilities where they're all in a row, so add those back in to get $420-60+6=366$. Subtract that from the total number $\binom{9}{3,3,3}=1680$ to get $1680-366=1314$.
You can use the principle of inclusion and exclusion. $$p - p_A - p_B - p_C + p_{A,B} + p_{A,C} + p_{B,C} - p_{A,B,C}$$ where there $p$s denote the number of permutations in which the letters of the subscript are together, i.e. $p_{A,B}$ denote the number of permutations in which the $A$s and the $B$s are grouped in three.
It is a bit tedious, but we end up with $$1680 - 3\times 140 + 3\times 20 - 6 = 1314$$ possible rearrangements.