The letter A can be assigned in 26 ways
The letter B can be assigned in 25 ways
The letter Z can be assigned in 1 ways
So the answer is 26!
and in Euler constant form is $e^{61.26170}$

However, the answer in the text book is $≈ (26!)^2/e$

Could you please help me with this problem?

  • 2
    $\begingroup$ Your $26!$ would not ensure all the code letters different to their original letters. $(26!)^2/e$ is even bigger so too large, though might have a typo. You might want to investigate derangements $\endgroup$
    – Henry
    Dec 1 '21 at 9:36
  • 1
    $\begingroup$ @coffeemath because $26!$ is squared in $(26!)^2/e$ $\endgroup$
    – Henry
    Dec 1 '21 at 12:20

I think the book is (implicitly or explicitly) looking for codes where a letter is never mapped onto itself. In other words, it is asking for the number of derangements of 26 objects, which is approximately equal to $\frac{26!}{e}$.

The exact formula is

$$n! \sum_{i=0}^n \frac{(-1)^i}{i!}$$

(which can be derived with the inclusion-exclusion principle)

but for $n \ge 1$ you can just round $\frac{n!}{e}$ to the nearest integer.


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