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?

Guest Jan 18, 2022

#1**0 **

The way I would do it is first take the outcomes when all digits are different. This would be 6*5*4 or 120 outcomes. Now we can do the rest of the outcomes. First, we can see the outcome 555. Then, we can do the outcomes when the number is 55_ with the _ being an unknown digit. This leads to 5 more outcomes. There are 5 more outcomes for 5_5 and 5 more for _55. Up till now we have gotten 136 outcomes. Now we can do the ones where there are 2 6's. So we have 66_, 6_6, and _66. We can see this is just like with 2 5's, so we get 15 more outcomes. Now, there are now more outcomes, and the answer is 151 different positive 3-digit integers.

I might've gotten this wrong, but I think I am correct.

ImDecentAtMath Jan 18, 2022