## College Algebra (10th Edition)

According to the Multiplication Principle, if we have $p$ selections for the first choice, $q$ selections for the second choice, then we have $p\cdot q$ different ways of selections and so on for more choices... Here we have $26-3=23$ possibilities for the third character, $26$ possibilities for the first two characters and $10$ possibilities for the four numbers, thus the number of all possibilities is $23\cdot26^2\cdot10^4=155480000$