Consider all the $6$ digit numbers that can be formed using the digits $1,2,3,4,5$ and $6$,each digit being used exactly once. Find number of such six digit numbers which have the property that for each digit,not more than two digits smaller than that digit appear to the right of that digit.
My attempt:
There appears to be an element of recursion. Assume that digits given are $1,2,3...,n-1,n$. Let us assume that there are $a_{n}$ of such n-digit numbers satisfying the above property. If we remove $n$ and then there are exactly $a_{n-1}$ such numbers.
So to create $n$ digit such number we have $3$ choices for $n$ to be put in $a_{n-1}$ such numbers.
(i) $n$ can be put to right of the $(n-1)$th digit.
(ii)$n$ can be put to right of the $(n-2)$th digit but to the left of $(n-1)$th digit.
(iii)$n$ can be put to right of the $(n-3)$th digit but to the left of $(n-2)$th digit.
Thus it can be said that $a_{n}=3a_{n-1}$
So,$a_{6}=3a_{5}=3^2a_{4}=3^3a_{3}=3^4a_{2}=3^4(2)=162$