In a four-letter code, each letter is A, B, C, or D. How many possible codes are there, if each letter can be used at most twice?

Guest Sep 22, 2020

#1**0 **

It is always much easier to list the combinations first and convert them to permutations:

{A, A, B, B} | {A, A, B, C} | {A, A, B, D} | {A, A, C, C} | {A, A, C, D} | {A, A, D, D} | {A, B, B, C} | {A, B, B, D} | {A, B, C, C} | {A, B, C, D} | {A, B, D, D} | {A, C, C, D} | {A, C, D, D} | {B, B, C, C} | {B, B, C, D} | {B, B, D, D} | {B, C, C, D} | {B, C, D, D} | {C, C, D, D} = 19 combinations.

Their permutations =[6 x 6] + [12 x 12] + 24 =36 + 144 + 24 =**204 permutations, or number of possible codes**

Guest Sep 22, 2020