Timeline for Young tableau with no i in row i, name that derangement
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 16, 2015 at 0:18 | answer | added | Richard Stanley | timeline score: 3 | |
| Apr 15, 2015 at 19:00 | comment | added | Zack Wolske | @Per: It's my understanding that Young tableaux are diagrams filled with anything, and standard Young tableaux are filled with $[n]$ in increasing order along rows and down columns. That may not be universal, and if it's confusing I can change it. The problem doesn't seem to me to be related to the usual representation topics that standard Young tableaux are used for, but if it is that would be good too. | |
| Apr 15, 2015 at 18:04 | comment | added | Per Alexandersson | It is not really a Young tableau, since the columns are not in increasing order, (right)? So maybe use diagram filling instead? | |
| Apr 15, 2015 at 17:51 | history | edited | Zack Wolske | CC BY-SA 3.0 |
added clarifying remarks per comments
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| Apr 15, 2015 at 17:49 | comment | added | Zack Wolske | @MaskedAvenger: No problem, thanks for reading it. It turned out to be more complicated to write down than I thought it would be. I'll make some clarifying edits. | |
| Apr 15, 2015 at 17:45 | comment | added | The Masked Avenger | I now saw your replying comment; thanks for bearing with my obtuseness. | |
| Apr 15, 2015 at 17:43 | comment | added | The Masked Avenger | Ok. The last thing I missed (I hope) is that the tuple of questions asked is an initial fragment of a permutation of the questions handed out. This makes it more of a challenge. I would suggest clarifying the point that no two students will be asked the same question, and that all permutations of (the order of) the n questions are equally likely to occur. | |
| Apr 15, 2015 at 17:40 | comment | added | Zack Wolske | @MaskedAvenger: questions asked are distinct, no two students will be asked the same one. So the $n-a$ available questions will decrease, or not, when one is asked, depending on whether a student worked on that question. | |
| Apr 15, 2015 at 17:27 | comment | added | The Masked Avenger | I see one thing I missed: a student can work on more than one question. I see also that there are more questions than students. However, if no students share a question, then we get a product of k terms of the form (n-a)/n, where a is allowed to vary over sizes of a certain partition of n into k parts. Or did I miss more? | |
| Apr 15, 2015 at 17:20 | comment | added | The Masked Avenger | Either you are over thinking this, or the actual problem is not well described by the group assignment scenario (or I am missing something). Why is the probability different from (1 -1/n)^k? | |
| Apr 15, 2015 at 16:59 | history | asked | Zack Wolske | CC BY-SA 3.0 |