There are 5 slots. Two of them have to be 5 and 15. Three of them have to be anything else. The way I set it up, I said that the first slot has a 1/10 slot of being either 5 or 15, and if that happens, the 2nd slot has a 1/19 chance of being the other number. So this includes the probability of getting 5 then 15 or 15 then 5. But of course they don't have to be in the first and second slots, they could be in any of the 5 slots. so we have to choose which two slots they will be in or 5C2.
So, we choose where the 5 and 15 go, but the way I set up the problem already includes the possibility of different orders of 5 and 15. And we don't have to worry about the different orders of the "anything else" slots because the calculation already includes all the different orders of those slots. For example, 5 15 7 8 9 and 5 15 9 8 7 are both already considered because when we say the three slots can be ANYTHING else, the last 3 could go 7 8 9 just as easily as 9 8 7.
