Problem: In the word 'combination' How many arrangements have no two adjacent letters the same?
My solution:
Using the inclusion-exclusion principle as follows:
$|A_1 \cup B_1 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2|- |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|$
$|A_1|=|A_2|=|A_3|= \frac{10!}{2!2!}$
$|A_1 \cap A_2|=|A_1 \cap A_3|=|A_2 \cap A_3| = \frac{9!}{2!}$
$|A_1 \cap A_2 \cap A_3| = 8!$
So to find the total arrangements where no two letters are adjacent I subtracted all the cases of adjacent letters from the total of arrangements with no restrictions, so I have:
$\frac{11!}{2!2!2!} - 3\times\frac{10!}{2!2!} - 3 \times \frac{9!}{2!} + 8!$
However I am aware this is wrong and I am supposed to add the $3 \times \frac{9!}{2!}$ instead, can't seem to wrap my head around why this is because why would we want to add instance where two adjacent letters occur?
Edit: I have found my issue thanks to the responses, I simply forgot that the sign flips to a + because of the two minuses, a mistake to remember for next time!