How many 3-digit numbers can you make with the digits 4, 7, and 8 if each digit can appear only once?

Answer 1

3! = 3x2x1 = 6 ways

478

487

748

784

847

874

nPr = n!/(n-r)! order matters, so you use permutations not combinations

3P3 = 3!/(3-3)! = 3!/0! = 3!/1 = 3! = 3x2x1 = 6