Given the digits 1, 2, 2, 3, 3, 3, how many 4-digit numbers can we build from these digits?
You have 5 distinct combinations. (1,2,2,3), (1,2,3,3), (1,3,3,3), (2,2,3,3), (2,3,3,3) Each combination has the following number of permutations: [4!/2! + 4!/2! + 4!/3! + 4!/2!2! + 4!/3!]=12 + 12 + 4 + 6 + 4 = 38 permutations.
