Let say that you have a set $${0,1,2}$$ and you pick a number 4 times from that set. Is there a function that gives me how many ordered outcomes can come from this?

Question

anonymous · Answer

I suppose

anonymous · Answer

I can think of algorithms that would give me the answer, but not functions. Do you have an example of the function?

anonymous · Answer

isn't it (n)!/(n-r)!

where n is the number of things to choose from and r is the number of things chosen

anonymous · Answer

@mattt9 It is part of a solution I think at least, But in the case I'm describing I would have to apply that function 15 times. Was hoping it would be possible with just one.

anonymous · Answer

Well what do you mean by ordered outcomes? Why would you have to do it 15 times?

anonymous · Answer

Well, I can choose 4 numbers, so I could get 0000, which is ordered 0000 and I can get 0001 which is ordered 0001,0010, 0100,1000 and i can get all combinations up to 2222. So 0000 to 2222 is 15 combinations (what I got at least). and all of them have to ordered. I would have to do (N!)/(n-r)!r! and depending on how many different numbers I have I have to do (N!)/(a!b!c!) where a+b+c equals N.