Alright, so we are dealing with permutations. Permutations are the number of combinations in a specific order possible for the set.
There's 4 letters, each one can be paired with 3 other letters, but those letters could be in different orders. To figure out how many variations of each combination there are (aka the number of permutations) use this formula:
[tex] _{n}P_{r}= \frac{n!}{(n-r)!} [/tex]
[tex]r=[/tex] number of elements in the subset
[tex]n=[/tex] number of elements in the set
[tex]P=[/tex] permutations of the set
There are only 3 elements in the subset because there is 1 that will not be repeated in each set, and there are 4 elements in the set.
Here's the math:
[tex] _{n}P_{r}= \frac{n!}{(n-r)!} [/tex]
[tex] _{4}P_{3}= \frac{4!}{(4-3)!}[/tex]
[tex] _{4}P_{3}= \frac{4!}{(1)!}[/tex]
[tex] _{4}P_{3}= \frac{4(3)(2)(1)}{1}[/tex]
[tex] _{4}P_{3}=24[/tex]
There are 24 permutations. I can prove this by showing you the model:
ABCD, ABDC, ACBD, ACDB, ADBC, ADCB are the 6 arrangements possible of the set starting with the letter A. Because there are 4 letters, the total amount of permutations without repeated letters is 4 (letters) times 6 (possible arrangements), which equals 24.
Hope this helps!
Answer: 24 passwords are possible
