I'm just a simple, uneducated guy with logic skills. To figure out how many possible combinations there are of anything, you lay out a slot for each possible number position in the series and then mutiply them all together. In your example there are 64 possibilities in each of 8 slots, therefore:
64x64x64x64x64x64x64x64 (This is assuming a number may occur multiple times in a single draw, as you showed above.)
If a number can only appear once in a draw, the number of possibilities would be:
64x63x62x61x60x59x58x57
As was already stated, the number of combinations are, um, staggering.
2007-02-23 16:46:12
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answer #1
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answered by vinny_the_hack 5
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I'm a little confused. Do you mean the digits 1-9, in combos of 8?
It would be 9 P 8, which is a different formula all together. 8! is 8*7*6*...
Ok... just looked it up... ther permutation formula is (n!)/(n - r)! That means 9!/(9-8)! = 9!/1! = 9! =362880... if it is 64, then it would be 64!/(64-8)! = 64!/56! = some big number... you can do it!
2007-02-23 23:25:52
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answer #2
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answered by Erynn 1
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2007-02-24 01:24:39
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answer #3
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answered by kolzig15 1
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8 factorial (8!)
2007-02-23 23:23:38
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answer #4
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answered by Kirk T 2
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Here is the answer. 9,223,372,036,854,775,808. All this other stuff is B.S. We wanted the answer, not the explanation!
2014-03-19 21:32:04
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answer #5
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answered by snfromky 1
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www.playwithnumbers.com
2007-02-27 14:38:00
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answer #6
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answered by Harry 5
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