Question

jose is choosing a 3-letter password from the letters a,b,c,d. the password cannot have the same letter repeated in it. how many passwords are possible

Answer 1

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:

$_{n}P_{r}= \frac{n!}{(n-r)!}$

$r=$ number of elements in the subset

$n=$ number of elements in the set

$P=$ 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:

$_{n}P_{r}= \frac{n!}{(n-r)!}$

$_{4}P_{3}= \frac{4!}{(4-3)!}$

$_{4}P_{3}= \frac{4!}{(1)!}$

$_{4}P_{3}= \frac{4(3)(2)(1)}{1}$

$_{4}P_{3}=24$

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