How many different positive three-digit integers can be formed using only the digits in the set {2, 3, 5, 5, 5, 6, 6, 7, 8} if no digit may be used more times than it appears in the given set of available digits?
since there are 9 numbers and choose 3 to be in so we can do 9!/6!=504 but some numbers are repeated in the list which are 5 and 6. 5 is repeated three times and 6 is reapeated twice so we divide by 2! and 3!
