You have an urn with 6 numbered balls and you pull out 3 (with replacement). What is the chance of getting 3 different numbers when the order doesn't matter?
I have two solutions that both seem reasonable.
Looking at chances:
Chance to pick the first ball: 6/6
Chance to pick the second ball: 5/6
Chance to pick the third ball: 4/6
i.e. 20/36Looking at possibilities (combinatorics):
To get 3 different balls out of a urn with 6 when no number can be repeated and the order doesn't matter is the same as a lottery.
Therefore there are ${6 \choose 3}$ = 20 possibilities.
Looking at all possibilities, I have 6*6*6 = 216. i.e. 20/216
Which one is wrong and why?