We have the word 'EXETASTIKH'. I want to calculate the number of rearrangements so that at least one E is at its original position.
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For that do we consider the first E constant and calculate the number of rearrangements of the remaining letters plus the same considering the second E constant?
Would it be then $\frac{10!}{2!}+\frac{10!}{2!}$ ?
As tdluong pointed out in a comment, apparently we both miscounted the letters. There are only $10$ letters in total, so only $9$ without the 'E's, so the answer is in fact
$$ \frac{9!}{2!}+\frac{9!}{2!}-\frac{8!}{2!}=342720\;. $$
I'm leaving the original, incorrect answer here, too, to make sense of the comments:
That's almost right, except you're double-counting the arrangements with both 'E's in their original positions, so you need to subtract their number, for a total of
$$ \frac{10!}{2!}+\frac{10!}{2!}-\frac{9!}{2!}=3447360\;. $$