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bonanova
Daedalian Member

 Posted: Fri Jul 17, 2009 8:45 am    Post subject: 1 I was playing with the following discrete puzzle and wondering if it has a continuous extension: n balls numbered 1-n are placed in each of 10 urns. A certain number m of balls is removed from each urn. The remaining balls have the following property: 1. if any 5 urns are emptied into a bowl, there will be at least one ball with each of the numbers 1-n 2. if any 4 urns are emptied into a bowl, there will be at least one missing number. The problem asks what is the smallest n that makes this possible, and what is the value of m? It does not ask which m balls are removed from the urns - a different subset is obviously removed from each urn. I reasoned that if 10 unique subsets of the set [1, ..., n] can be found with these two conditions for a finite n, there should be 10 unique subsets of the reals on [0, 1] with the same properties: the union of any 5 subsets covers the interval, but of any 4 does not. Maybe it's trivial, but I couldn't find a starting point. Any thoughts? OK all i had to do is post this and I see the answer. So let's add the question: what parts of the interval are removed in the ten cases?_________________ Vidi, vici, veni.
ralphmerridew
Daedalian Member

 Posted: Fri Jul 17, 2009 1:46 pm    Post subject: 2 Minimal n is 10C6 = 210. If any ball appears in 5 or fewer urns, then choose the other 5 urns and you'll miss that ball. If any ball appears in 7 or more urns, remove that ball and the result will have fewer balls but still have properties 1 and 2. Therefore every ball appears in exactly 6 urns. Consider any set of six urns. There must be at least one ball that appears in exactly those 6 urns, or choosing the other 4 would get all the balls. There must be at most one, or one of them could be removed without disturbing properties 1 or 2. Therefore, n = 10C6 in a minimal solution.
Zag
Tired of his old title

 Posted: Fri Jul 17, 2009 2:27 pm    Post subject: 3 I was going to say 10C4, for exactly the same reason (which is, of course, the same answer). My thinking was that for every group of 4 urns, there is exactly 1 ball missing, which must be in all the remaining urns.
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