Originally posted by: DeafeningSilence
Originally posted by: dman6666
I believe that is for ALL combinations. In a lottery analogy, there would be 90 balls numbered 0..9, and 9 slots.Originally posted by: DeafeningSilence
Originally posted by: dman6666
My reasoning:
9 Digits times 9 locations = 81 possible. (you can actually rule a few of these out for SS#'s because they haven't gotten up past 7 for the first digit, so that takes away a few 'real' combinations). But assuming the maximum...
And there are 10 digits * 81 (possible combinations with 9 spots) 81 * 10 = 810.
I could be very wrong. [edit]I don't think so though. I think some of the 9 voters lied. Otherwise we'd have a nice bell curve and we don't! Take that![/edit]
It's been a while since I had stats too. But wouldn't it be this? 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 = 3628800.
In the case I'm talking about, there are 10 balls (0..9) and 9 slots... so you can't have any repeating #'s.
123456789
012345678
901234567
890123456
789012345
678901234
567890123
456789012
345678901
234567890
are the 10 possible unique number sets. And each of those sets can be arranged 9x9 times = 81. So The total should be 10x81 = 810. And like the case where the folks picked #1 too many times, I think 10 people having non-repeating #'s in their ss# is an unusually large amount to be represented on ATOT. Based on the curve, at least 3 people lied so far.
I actually don't think my solution was all possible combinations. To get all possible combinations (duplication allowed), it would be 10x10x10x10x10x10x10x10x10 = 10^9. In my equation, there are 10 choices for the first digit, leaving 9 choices for the second digit, 8 for the third, and so forth. This would prevent duplication of any number.
10^9 is correct.
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