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S Sep 29, 2013 at 22:27 history suggested Yuichiro Fujiwara CC BY-SA 3.0
Fixed typo in author's name, added tags, and formatted title in latex
Sep 29, 2013 at 22:16 review Suggested edits
S Sep 29, 2013 at 22:27
Sep 29, 2013 at 8:31 answer added Yuichiro Fujiwara timeline score: 2
Apr 16, 2010 at 13:39 comment added Hung Q. Ngo Thanks Gerhard & Tony for the comments. @Gerhard, I did implicitly mention the $\binom{n-t}{k-t}$ bound in the last paragraph. @Tony, I'm aware of the Fisher's inequality. It's perhaps a little too far from what I'm seeking.
Apr 16, 2010 at 4:32 comment added Tony Huynh Somewhat related is Fisher's inquality. That is, if we insist that every two sets have intersection exactly $t$, then there can be at most $n$ such sets (where $n$ is the size of the underlying set). Indeed, this holds even in the non-uniform case.
Apr 16, 2010 at 0:59 comment added Gerhard Paseman Now that the math has rendered correctly, it seems the lower bound (n-t) choose (k-t) figures often, and is applicable when s <= (n-t) choose (k-t), and may not apply otherwise.
Apr 16, 2010 at 0:40 comment added Gerhard Paseman It might help to mention an obvious lower bound: (n-t) choose (k-t). Gerhard "Ask Me About System Design" Paseman, 2010.04.15
Apr 16, 2010 at 0:19 comment added Harry Gindi I've removed the LaTeX in the title because it does not render properly on the front page or the questions page.
Apr 16, 2010 at 0:18 history edited Harry Gindi CC BY-SA 2.5
delatexified the title
Apr 16, 2010 at 0:17 history asked Hung Q. Ngo CC BY-SA 2.5