prob.

I'm not sure if I'm undertanding the question properly. I think you're saying that a number has four digits and the only possible set of numbers that you can use for the digits are {3, 4, 5}. So 4 and 5 can be used only once and 3 can be used two times. Since there can be at most two 3's, then it must mean that 4 and 5 MUST be used exactly once each in the sequence and three MUST be used twice since there needs to be a total of 4 digits. So there are 4C2 ways to pick where the 3's go. There are 6 ways to order those 3's because 4C2 = 4!/(2!2!) = 12/2. In each of those 6 ways, there are 2 ways to order the 4 and 5. Either 4 before the 5 in the sequence, or vice versa. So there are 6*2 ways of ordering those numbers to make this phone number.

1. 3345
2. 3354
3. 4335
4. 5334
5. 4533
6. 5433
7. 3453
8. 3543
9. 3435
10. 3534
11. 4353
12. 5343

This question basically asks how many ways can we arrange the numbers 3345

To find the number of permutations of n objects of which r are duplicates, calculate n!/r!

4!/2! = 12