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I am trying to figure out how to calculate the number of ways to arrange a consecutive string of letters with no letter in the alphabetically correct place. For example, if we take $ABC$, then there are 6 possible arrangements:

$$ABC, ACB, BAC, BCA, CAB, CBA$$

but only in the last two none of the letters is in the alphabetically correct place (for example, $BAC$ does not satisfy the requirement, since $C$ is a third letter of the alphabet and it is in the third spot in $BAC$)

If we take $ABCD$ there are $4!=24$ possible arrangements, but by inspection one can see that only $9$ of the arrangements satisfy the given requirement.

My question is: what's the formula for calculating the number of ways to arrange a consecutive string of letters with no letter in the alphabetically correct place and how to deduce such a formula?

In fact, I am also trying to understand the following: why does the ratio of all the possible arrangements to the arrangements satisfying the extra condition converge to the famous Euler's number $e$?

As always, any help will be greatly appreciated.

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Just as Jose Carlos Santos posted in the comments, this can be solved by counting the number of derangements.

In particular a useful formula would be:

$$!n = \left[ \frac{n!}{e} \right]$$

where $[x]$ is the nearest integer function. So the probability you are looking for would be:

$$!n =\frac{1}{n!} \left[ \frac{n!}{e} \right] \rightarrow \frac{1}{e}$$

answering your second question

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