Please refer to the setless set problem on this page. AnalysisIt can be easily deduced that the collection of 81 cards, contiains 27 SETs where each set contains the same cards. Now in order to find the number of ways in which we can collect 20 cars such that it contains no set, we can use the following formula: $$ N_{20\,cards\,without\,set} = N_{all\,possible\,collections\,of\,20\,cards}  N_{collection\,of\,20\,cards\,with\,atleast\,1\,set}$$ Find out set with different cards Now if \(n_{same}\) is the number of sets with same cards, we already know that $$n_{same} = 27$$ If we want to form a set that contains different cards, we can first pick an equal set and then pick a card from that set. Repeating this thrice will give us a set that contains different cards. Thus total number of sets with different cards \(n_{different}\) can be defined as $$n_{different} = (27 * 3) * (26 * 3) * (25 * 3) $$ Thus total number of possible sets \(n_{sets}\) can be defined as: $$n_{sets} = n_{same} + n_{different}$$ Now we can easily find out the number of ways to pick k sets from \(n_{sets}\). A collection of 20 cards will contain a minimum of 1 set and a maximum of 6 sets. Thus number of ways to collect 'at the most' 6 sets can be defined as $$n_{6\,sets} = \sum_{k = 1}^{k = 6} \binom {n_{sets}} {k}$$ Thus to find out N ways in which we can collect 20 cards that does not contain a set is $$ N = \binom {81} {20}  n_{6\,sets}$$ ImplementationWe can now write code that
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MeI am a polyglot software engineer specializing in shipping iOS and 3d scientific visualization applications. Archives
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