Timeline for Number of A Subset of Monomials
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 10, 2010 at 12:50 | vote | accept | Joe Johnson | ||
| Jun 10, 2010 at 12:50 | vote | accept | Joe Johnson | ||
| Jun 10, 2010 at 12:50 | |||||
| Jun 9, 2010 at 23:58 | comment | added | JBL | Equivalently, replace the upper limit of summation with $\min(k, \lfloor n/2\rfloor)$. | |
| Jun 9, 2010 at 23:23 | comment | added | Steve Huntsman | Also I guess this is the same convention as Knuth and "A=B" use... | |
| Jun 9, 2010 at 23:20 | comment | added | Steve Huntsman | OK, this looks good: for $k=4$ and $n=3$ this gives $\binom{6}{3}-\binom{4}{1}\binom{1}{1}-\binom{4}{2}\binom{0}{1}-\binom{4}{3}\binom{-1}{2}-\binom{4}{4}\binom{-2}{3}$, which would be interpreted as $\binom{6}{3}-\binom{4}{1}\binom{1}{1} = 20-4 = 16$, which checks out. | |
| Jun 9, 2010 at 22:55 | comment | added | Vladimir Dotsenko | Well, the convention I would definitely like to follow is $\binom{p}{q}\ne 0$ only for $0\le q\le p$, and in this case is given by the usual formula. | |
| Jun 9, 2010 at 22:12 | comment | added | Steve Huntsman | I don't know how to make sense of this if $k > n$. | |
| Jun 9, 2010 at 20:15 | history | answered | Vladimir Dotsenko | CC BY-SA 2.5 |