We know that the Monoalphabetic Substitution Cipher should have a lot more keys than the Caesar Cipher, but how many more?

Ptext: ABCDEFGHIJKLMNOPQRSTUVWXYZ
Ctext: ??????????????????????????

Let's look at the first "?" under the "A" in the ciphertext.
How many choices do we have for that "?". Well, we have choices because we can choose any letter of the alphabet. Let us say we chose "R".

Ptext: ABCDEFGHIJKLMNOPQRSTUVWXYZ
Ctext: R?????????????????????????

How many choices do we have for the next "?". We can choose any letter of the alphabet except "R", because we have already chosen that for the plaintext letter "A". So we have choices. Let's say we chose "E". Now for the next question mark we can't chose the letter "R" or "E" so we have choices. By now you should see the pattern, we have choices. That is equivalent to factorial.

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This is a lot of keys. Much much more keys than a Caesar Cipher. Unfortunately, with today's computers this is not that many keys. But this makes it impossible to try to crack a monoalphabetic substitution cipher using brute force by hand. There is however another method to crack this code...