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Aug 11, 2017 at 14:19 history edited Felix Torssell CC BY-SA 3.0
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Aug 11, 2017 at 14:18 vote accept Felix Torssell
Aug 10, 2017 at 21:58 answer added Stingy timeline score: 1
Aug 10, 2017 at 16:13 answer added toto2 timeline score: 1
Aug 10, 2017 at 8:59 answer added Felix Torssell timeline score: 0
Aug 10, 2017 at 7:37 answer added Vogel612 timeline score: 0
Aug 10, 2017 at 7:29 history edited Felix Torssell CC BY-SA 3.0
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Aug 10, 2017 at 7:28 comment added Felix Torssell I took 325 as n because that is my current number of options. sorry I should have clarified. I edited the question
Aug 10, 2017 at 7:24 history edited Vogel612
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Aug 10, 2017 at 7:16 comment added Stingy \$n\choose r\$ is equivalent to \$\frac{n!}{r!\cdot(n-r)!}\$. It doesn't count permutations. So the question how you got 5668650 still stands.
Aug 10, 2017 at 5:48 comment added Felix Torssell That is correct Stingy. Each serial number has about 60 lines. 1 for each of its options. And it can't have the same option twice.
Aug 10, 2017 at 5:46 comment added Felix Torssell I don't want permutations. I am interested in combinations and I don't want position specific or repetition. Formula I am using is n!/r!(n-r)!
Aug 9, 2017 at 21:43 comment added Stingy Also, I am confused how you got the number \$5668650\$, could you elaborate on that please? \${350\choose 3}=7084700\$, and \${350\choose 2}=61075\$, and \$2^{350}=2.29349861599007 \cdot 10^{105}\$, so maybe I misunderstood something in your question.
Aug 9, 2017 at 20:41 comment added Stingy Just to clarify: Each serial number can have an arbitrary number of options associated with it, and it's just a characteristic of the input format that a line contains just one option for a serial number?
Aug 9, 2017 at 11:55 review First posts
Aug 9, 2017 at 12:13
Aug 9, 2017 at 11:53 history asked Felix Torssell CC BY-SA 3.0