Timeline for Permutation count for sets with multiplicities
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 13, 2014 at 16:11 | answer | added | Mr.Wizard | timeline score: 2 | |
| Dec 26, 2013 at 13:59 | history | edited | Mr.Wizard |
edited tags
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| Dec 26, 2013 at 13:12 | comment | added | ubpdqn | @Nasser I agree that is unclear how the word 'group' is to be interpreted. The first answer counted permutations. I merely presented another approach for that objective. OP will presumably clarify. | |
| Dec 26, 2013 at 10:45 | history | tweeted | twitter.com/#!/StackMma/status/416157793629257728 | ||
| Dec 26, 2013 at 8:39 | answer | added | ubpdqn | timeline score: 8 | |
| Dec 26, 2013 at 8:37 | history | edited | m_goldberg | CC BY-SA 3.0 |
General cleanup
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| Dec 26, 2013 at 7:38 | review | Close votes | |||
| Dec 26, 2013 at 14:25 | |||||
| Dec 26, 2013 at 6:40 | answer | added | VF1 | timeline score: 6 | |
| Dec 26, 2013 at 6:22 | comment | added | Nasser | I must be missing the question. But isn't the length of the list the same as number of groups? Length@Permutations[{a, a, b, c, c, d}] gives 180. Or you looking for some other number? | |
| Dec 26, 2013 at 5:57 | review | First posts | |||
| Dec 26, 2013 at 7:22 | |||||
| Dec 26, 2013 at 5:42 | history | asked | Dhruba Roy | CC BY-SA 3.0 |