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Find the permutation of $\{1,2,3,4,5,6\}$ such that the pattern $13$ and $246$ do not appear.

What I did in this question is found number of permutation of the $6$ given number I treated $13246$ as a single number and found the number of permutation of these $5$ numbers and subtracted $5!$ from $6!$ to get $600$ from here I am not able to proceed further.

Any help would be appreciated. Thanks for advance.

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The number of permutations in which $13$ appears is $5\times 4!$.

The number of permutations in which $246$ appears is $4\times 3!$.

The number of permutations in which $13$ and $246$ both appears is $6$, namely

$132465, 135246, 513246, 246513, 524613,264135$

So by exclusion-inclusion your ans is $6!-5\times 4!-4\times 3!+6=582$.

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  • $\begingroup$ But the number of permutations in which both $13$ and $246$ appear is simply $3!=6$, NOT $5$. You're missing $246135$. $\endgroup$ – zipirovich May 24 '17 at 16:04
  • $\begingroup$ There is something wrong because the answer is 582. $\endgroup$ – user449276 May 24 '17 at 16:08
  • $\begingroup$ There are six permutations in which $13$ and $246$ both appear. You omitted $26135$. Notice that we have three objects to permute, $13$, $246$, and $5$. They can be arranged in $3! = 6$ ways. $\endgroup$ – N. F. Taussig May 24 '17 at 16:46
  • $\begingroup$ Corrected it, now its fully correct $\endgroup$ – Arpan1729 May 24 '17 at 16:48

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