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Find the number of permutation of {1,2,3,4,5,6} such that the patterns 13 and 246 do not appear. Show the steps .

apnorton
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U-571
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    This is not a "do my homework for me" site. Explain us what you've tried. – leonbloy Nov 27 '13 at 03:26
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    For the future, this sort of question is not about [tag:permutation-groups]... That's a different topic. ;) – apnorton Nov 27 '13 at 03:28
  • Thank you @leonbloy for reminding ! but my intension was not that. I have solved it and wanted to check my solutions is correct or not. – U-571 Nov 27 '13 at 05:00
  • @U-47 Then you should post your proposed solution as part of your question. It is easier to verify that a solution is correct than to find a solution from scratch. (Besides, what if you get an answer with a solution that is different from yours? That doesn't mean that yours is incorrect.) – Trevor Wilson Nov 27 '13 at 05:04

1 Answers1

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It is easier to first find the number of permutations in which the pattern $13$ or $246$ does appear.

For the pattern $13$, tie $1$ and $3$ together to make a supersymbol. Together with $2,4,5,6$, and the supersymbol. we have $5$ "symbols" that can be arranged in $5!$ ways.

Similarly, there are $4!$ ways in which the pattern $246$ appears.

There are $3!$ ways in which both patterns appear. These are counted twice in the sum $5!+4!$.

Thus there are $5!+4!-3!$ ways thast one or both of our patterns can appear. It follows that there are $6!-(5!+4!-3!)$ ways to arrange our symbols so that neither pattern appears.

apnorton
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André Nicolas
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