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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

User Donmartin
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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)= (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)= (n!)/((n-r)!)


_(4)P_(3)= (4!)/((4-3)!)


_(4)P_(3)= (4!)/((1)!)


_(4)P_(3)= (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
User Anthony Alberto
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