IMO 1968 SL 24

Origin: MON

Problem

Find the number of all $n$-digit numbers for which some fixed digit stands only in the $i$th $(1 < i < n)$ place and the last $j$ digits are distinct.$^3$

Solution

If the $i$th digit is $0$, then the result is $\frac{9^{k-j}9!}{(10-j)!}$ if $i > k-j$ and $\frac{9^{k-j-1}9!}{(9-j)!}$ otherwise. If the $i$th digit is not $0$, then the above results are multiplied by $8$.