>>10881654If I have 10 balls labeled from 0-9, how many ways can I arrange them?
A good way to think about it is to consider how many choices I have for the first ball, then the second, then the third, etc. So I will have 10 choices at first, then 9, then 8, then 7 until there is only one ball remaining. So I have 10*9*8*7*6*5*4*3*2*1 total possible arrangements that can be made.
Another thing to note is that all arrangements are equally likely, so the likelihood of getting the sequence 2, 3, 7, 6, 5, 4, 1, 8, 9, 0 is the same as the likelihood of getting 3, 7, 6, 5, 4, 1, 8, 9, 0, 2. Keeping this in mind and the fact that all of the possibilities have to add up to 1, the chance of getting any particular sequence in this scenario is 1/(10!).
This is your basic thought experiment that you will need to extend to different scenarios. For instance, what are the odds of getting a particular sequence if I only choose 5/10 of the balls? What happens if I don't care about how the numbers are ordered but only whether I got that number? What happens when I make some of the balls indistinguishable by only letting the balls have the number 1 or 2?
Considering these things will lead to the definitions of permutations, combinations, and the binomial theorem respectively. Teaching prob and stats in a 4chan thread is a bit difficult and I'm not getting paid for it, so you'll have to do the rest yourself. Look at Khanacademy, he's pretty good about spelling things out.
