Question

All the $7$ digit numbers containing each of the digits $1,2,3,4,5,6,7$ exactly once, and not divisible by $5$ are arranged in the increasing order. Find the $(2000{)}^{th}$ number in this list.

Answer 1

The no. of $7$ digit no. with $1$ in the left most place and containing each of the digit $1,2,3\dots 7$ exactly is $6!=720$,

But $120$ of these end in $5$ and hence are divisible by $5$. There the no. of $7$ digit no. with $1$ in the left most place and containing each of the digit containing $1,2,3\dots 7$ exactly once but no divisible by $5$ is $600.$

Similarly the no. of $7$ digit no. with $2$ & $3$ in the left most place and containing each of the digit $1,2,3\dots 7$ exactly once but not divisible by $5$ is also $600$ each.

Hence ${2000}^{th}$ no. must have $4$ in the left most place. Again the no. of such $7$ digit no. begin with $41,42$ and not divisible by $5$ is $120-24=96$ each and the account for $192$ no.'s.

These show that, ${2000}^{th}$ no. must begin with $43$ and the next $8$ no. in the list are, $4312567,4312576,4312657,4312756,4315267,4315276,4315627,4315672.$