Timeline for Taking Seats on a Plane
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2023 at 9:52 | comment | added | Kapes Mate | Oh...nope...the empty set must be counted as well... | |
| Feb 2, 2023 at 9:31 | comment | added | Kapes Mate | In this formula: $\sum_{J\subseteq I}\prod_{j\in J}x_{j}=\prod_{i\in I}\left(1+x_{i}\right)$, shouldn't we write instead $\sum_{J\subseteq I}\prod_{j\in J}x_{j}=-1+\prod_{i\in I}\left(1+x_{i}\right)$? | |
| Aug 28, 2021 at 4:56 | comment | added | user53259 | math.stackexchange.com/q/4230917 details this answer by Bryon Schmuland. | |
| Jun 8, 2019 at 7:39 | history | edited | Hans | CC BY-SA 4.0 |
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| Aug 19, 2014 at 2:25 | comment | added | Matt | See my comment in my answer for an intuitive explanation of your general formula. | |
| Aug 19, 2014 at 2:09 | comment | added | Matt | +1 for the reference! here's a link to it | |
| Jun 10, 2014 at 16:09 | history | edited | user940 | CC BY-SA 3.0 |
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| Jun 7, 2014 at 23:52 | history | edited | user940 | CC BY-SA 3.0 |
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| Oct 1, 2013 at 15:05 | comment | added | athos | thank you for your explanation. for point 1, after drawing it out i finally understand it. but for point 2, could you please elaborate a bit more? scenario A: the first person in the line bumped into seat #1, customer #1 then bumped into seat #5, this is $1\longrightarrow j_1=5\longrightarrow j_2\longrightarrow\cdots\longrightarrow j_m\longrightarrow k$; Scenario B: the first person in the line bumped into seat #5, customer #5 then bumped on, this is $j_1=5\longrightarrow j_2\longrightarrow\cdots\longrightarrow j_m\longrightarrow k$ -- these are 2 different scenarios right? | |
| Oct 1, 2013 at 12:14 | comment | added | user940 | 2. No, any bumping can be traced back to passenger 1. | |
| Oct 1, 2013 at 12:13 | comment | added | user940 | 1. $\prod_{i\in I}(1+x_i)=\sum_{J\subseteq I}\prod_{j\in J}x_j$ | |
| Oct 1, 2013 at 1:18 | comment | added | athos | may i ask 2 questions, 1: how could I get $\sum_{J\subseteq\{2,\dots,k-1\}} \prod_{j\in J}{1\over (n+1)-j} = \prod_{j=2}^{k-1} \left(1+{1\over (n+1)-j}\right)$? 2. the bumping may not start from customer 1, it could start from anyone. e.g. the diagram could be $5\longrightarrow j_1\longrightarrow j_2\longrightarrow\cdots\longrightarrow j_m\longrightarrow k$ if the first person in the line (lost his ticket) seats at seat #5. right? | |
| Aug 7, 2011 at 12:30 | history | answered | user940 | CC BY-SA 3.0 |