momo1729 wrote: Problem 3 (MAR CP 1992) : From the digits $1,2,...,9$, we write all the numbers formed by these nine digits (the nine digits are all distinct), and we order them in increasing order as follows : $123456789$, $123456798$, ..., $987654321$. What is the $100000th$ number ? The first $8!=40320$ such numbers are all these numbers beginning with $1$ The next $8!=40320$ such numbers are all these numbers beginning with $2$ So the $100000^{th}$ such number is the $100000-2\times 40320=19360^{th}$ such number beginning by $3$ The first $7!=5040$ such new numbers are all these numbers beginning with $31$ The next $7!=5040$ such new numbers are all these numbers beginning with $32$ The next $7!=5040$ such new numbers are all these numbers beginning with $34$ So the $19360^{th}$ such new number is the $19360-3\times 5040=4240^{th}$ such number beginning by $35$ The first $6!=720$ such new numbers are all these numbers beginning with $351$ The next $6!=720$ such new numbers are all these numbers beginning with $352$ The next $6!=720$ such new numbers are all these numbers beginning with $354$ The next $6!=720$ such new numbers are all these numbers beginning with $356$ The next $6!=720$ such new numbers are all these numbers beginning with $357$ So the $4240^{th}$ such new number is the $4240-5\times 720=640^{th}$ such number beginning by $358$ The first $5!=120$ such new numbers are all these numbers beginning with $3581$ The next $5!=120$ such new numbers are all these numbers beginning with $3582$ The next $5!=120$ such new numbers are all these numbers beginning with $3584$ The next $5!=120$ such new numbers are all these numbers beginning with $3586$ The next $5!=120$ such new numbers are all these numbers beginning with $3587$ So the $640^{th}$ such new number is the $640-5\times 120=40^{th}$ such number beginning by $3589$ The first $4!=24$ such new numbers are all these numbers beginning with $35891$ So the $40^{th}$ such new number is the $40-1\times 24=16^{th}$ such number beginning by $35892$ The first $3!=6$ such new numbers are all these numbers beginning with $358921$ The next $3!=6$ such new numbers are all these numbers beginning with $358924$ So the $16^{th}$ such new number is the $16-2\times 6=4^{th}$ such number beginning by $358926$ And so $358926-147$, $358926-174$, $358926-417$, $358926-471$ Hence the answer : $\boxed{358926471}$

You rock again, dear pco!