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.
If you're just interested in the result: Brute force tells me $3513600$.

But let's figure this out. There are $10!$ permutations in total. We have to subtract those with $ABE$. There are $8!$ of them (just treat $ABE$ as a single letter). Likewise, we have $8!$ for each of the other two three letter patterns. That leaves us with $10!-3\cdot8!$. Now we have subtracted patterns like $\ldots{}ABE\ldots{}HID\ldots$ twice. How many of them are there? The number of letters left is $4$ with $4!$ permutations of them and exactly two out of 5 spots (the spots before between and after the four letters) where the two patterns are inserted. Assume the first pattern goes into the first slot. Then the second one can also go there or into the second or third and so on. If the first pattern goes into the second slot the second one can only go into the second or third and so on. This means there are $\sum\_{i=1}^{5}i=15$ possibilities. As $ABE$ can be the first or the second pattern, we multiply by $2$ which gives us a total of $4!\cdot15\cdot2=720$ permutations with both $ABE$ and $HID$ in them. Now, $HID$ and $BED$ can obviously never occur simultaneously. But $ABE$ and $BED$ can. How often does that happen. Just treat $ABED$ as one letter then there are $7!$ of them. They also need to be put back in. We end up with $$10!-3\cdot8!+4!\cdot15\cdot2+7!=3513600$$