Timeline for probability of non-existence of a sum subset
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Dec 2, 2013 at 21:39 | comment | added | Aaron Meyerowitz | I was counting out of (unordered) sets of 3 values. so $0,x,y$ occurs $\frac{14\cdot 13}{2}=91$ ways. $x,-x,y$ with $y \ne 0$ happens $14\cdot 6=84$ ways. Another $9$ are $x,y,-x-y$ for 1,2 1,3 1,4 1,5 1,6 2,3 2,4 2,5 3,4 and a final $9$ of $-x,-y,x+y$ for the same choices . In all, $91+84+9+9=193.$ | |
| Dec 2, 2013 at 9:05 | comment | added | Turbo | are your counts correct? For n=3. In [a1 a2 a3], pick a1=0, then a2 and a3 can be any of the other numbers for a count of 14*13. Pick a1=1. Then you have a2=-a3 or a2=-1 or a3=-1. For a2=-1 itself you have 13 different choices for a3. This partical count itself takes number of ways 0 can occur as subset to 14*13+13=195>193. | |
| Dec 1, 2013 at 23:40 | history | undeleted | Aaron Meyerowitz | ||
| Dec 1, 2013 at 23:40 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
added 8 characters in body
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| Dec 1, 2013 at 23:06 | history | deleted | Aaron Meyerowitz | via Vote | |
| Dec 1, 2013 at 23:05 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |