# Number of permutation of natural number

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.

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 $6!-5\times 4!-4\times 3!+6=582$.

• But the number of permutations in which both $13$ and $246$ appear is simply $3!=6$, NOT $5$. You're missing $246135$. – zipirovich May 24 '17 at 16:04
• There is something wrong because the answer is 582. – user449276 May 24 '17 at 16:08
• 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. – N. F. Taussig May 24 '17 at 16:46
• Corrected it, now its fully correct – Arpan1729 May 24 '17 at 16:48