Question 278532
Let's use AAB.

3P3 = 3!/(3 - 3)!

3P3 = 6/0!

3P3 = 6/1

3P3 = 6 different PIN codes are possible.

Or you can do it the long way.

AAB becomes AA, AB, BA = 2 letters + 2 letters + 2 letters = 6 letters or PINS.