Timeline for Minimum differences in vectors of naturals
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Aug 13, 2010 at 3:19 | history | edited | Peter Shor | CC BY-SA 2.5 |
updated to discuss vectors and not sets
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| Aug 13, 2010 at 0:28 | vote | accept | Robby McKilliam | ||
| Aug 13, 2010 at 0:28 | comment | added | Robby McKilliam | Indeed, I mean vectors. I have fixed the title. I have made a bit of a mess of this question! And multiplying Peters answer by $m!$ as you have said give the desired result. | |
| Aug 12, 2010 at 23:44 | comment | added | Aaron Meyerowitz | It seems that you really do mean vectors and not multisets as in your amended title. But the answer for multisets is cute. Then $N_k$ is as Peter said for $k \gt 0$ and $N_0=\binom{n+m-1}{m}-\binom{n}{m}$ which is the $k=0$ case of the formula. | |
| Aug 12, 2010 at 23:18 | comment | added | Aaron Meyerowitz | That is the answer for sets since one gets just those sets, and once each. Then $\binom {n}{m} = \sum_{k=1}^{n-1}N_k$. If you want to have an $N_0$ and $n^m = \sum_{k=0}^{n-1}N_k$ then you really are talking about vectors. Then, for $k \gt 0$, multiply the answer above by $m!$. And then $N_0$ is what it would have to be, $n^m-\frac{n!}{(n-m)!}$ | |
| Aug 12, 2010 at 22:14 | comment | added | Robby McKilliam | My comment makes no sense! Repetition implies that $d(S) = 0$. I'm going to have to stop and think about what I am asking. | |
| Aug 12, 2010 at 21:57 | comment | added | Robby McKilliam | Hi Peter, I don't think this is correct. With ''m-subsets of $\{1,2,\cdots,n\}$'' you are not allowing repetition, but I mean for the question to allow repetition. This is wholly my fault, I should never have called $S$ a set! | |
| Aug 12, 2010 at 13:38 | history | answered | Peter Shor | CC BY-SA 2.5 |