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Let us say we have 52 cards with values ranging from 1-13 (4 sets of cards from 1-13). Assume that you wanted no two consecutive values to be next to each other in the pile of cards. For example, a 3 cannot be next to a 2 or a 4. How many ways can I arrange these cards that there are no consecutive values next to each other?

Can someone suggest a permutation that fulfills these requirements or suggest a computer program to solve the problem?

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    $\begingroup$ I do not believe this is a trivial problem. The answer will come from some integral of $e^{-x}$ type of function. Would be interested to see an easier solution. ! $\endgroup$ – Shailesh Oct 24 '15 at 16:38
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total ways are 52!. now let us assume we have two cards consecutive . consider them as a single group we can have 104 ways . they can be arranged in 2! ways between themselves.We have 1 set of 13 so number of ways of 1 pair of two consecutive cards is 2!.11! such arrangements in 104 ways can be done and of four different sets and these four sets can be arranged in 4! ways. The total ways that cards are not together =52!-(2!.11!.4!.104.4)

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    $\begingroup$ the total number of ways is $(52!)/(4*4!)$ though not 52! $\endgroup$ – Jaywalker Oct 22 '15 at 15:14
  • $\begingroup$ $(52!)/(13*4!)$ sry $\endgroup$ – Jaywalker Oct 23 '15 at 7:04
  • $\begingroup$ But you havent mentioned whether all numbers in 4 set are identical . Like 1 of some set x can also be placed in y. The probability changes for identical and distinct cards . If u mention that the answers you will get will be more precise . Thanks $\endgroup$ – Archis Welankar Oct 23 '15 at 7:34
  • $\begingroup$ Please see my comment I have placed after the question. Also, the total number of ways should be $52!/(4!)^{13}$ $\endgroup$ – Shailesh Oct 24 '15 at 16:40

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