Question: > 
> Count the number of arrangements of the $10$ letters ABCDEFGHIJ in which none of the patterns ABE, BED, or HID occur.
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I thought that the answer would be $\_{10}P\_{10} - (\_{10}P\_{3} + \_{10}P\_{3} + \_{10}P\_{3} - \_{10}P\_{2}$), since we have $\_{10}P\_{10}$ total arrangements, and $\_{10}P\_{3}$ arrangements for ABE, BED, and HID. But we subtracted $BE$ twice, so we need to add it back in: $\_{10}P\_{2}$.

Please tell me what is wrong with my reasoning!

Answer: 
$\_{10}P\_3$ is not the number of arrangements where $ABE$ occurs. You have $7!$ ways to put the other letters in order and $8$ places to put the unit $ABE$ so $8!$ ways to put all the letters in order including $ABE$. Alternately, you can group $ABE$ into one letter, have $8$ letters, which again gives $8!$ orders. 

Your double subtraction is combinations with $ABED$ and combinations with both $ABE$ and $HID$, so you need to add those back in once.
Since $ABE$ and $HID$ are disjoint, all Permutations where both patterns occur are subtracted twice.