Yes, there does exist such a way. First you select a digit d from {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Then you select a digit e from ({0, 1, 2, 3, 4, 5, 6, 7, 8, 9}-d). Then you select a digit f from (({0, 1, 2, 3, 4, 5, 6, 7, 8, 9}-d)-e). You first select 0 for d, then 1, and so on until you get to 7. And, you always select the least digit first for e and f also, with the additional condition that d < e < f. List out the first sequence, 012, 013, 014, 015, 016, 017, 018, 019. Then list all the other numbers beneath them with the condition that for all numbers e and f, and with d held constant, the digits for e and f follow the natural number sequence down the column. Partition each set of sequences by d. The column rule only applies within each partition. (this description might come as incomplete or could use some revision).
The list thus goes:
012, 013, ..., 019
023, 024, ..., 029
034, 035, ..., 039
.
.
.
089
123, 124, ..., 129
134, 135, ..., 139
.
.
.
189
.
.
.
789
I'll clarify the last part here:
567, 568, 569
578, 579
589
678, 679
689
789
Perhaps better, say we try to do the same thing in base 4. The entire sequence goes
012, 013
023
123
In base 5 the process yields:
012, 013, 014
023, 024
034
123, 124
134
234
Thus, in base 10 the sum of the first 8 triangular numbers gives us the number of such combinations: +(1, 3, 6, 10, 15, 21, 28, 36)=120. Similarly, it should logically follow that for x digit numbers in base z, where x < z, or x=z, there exist +[T$_1$, ..., T$_ (z-(x+1))$] such combinations, where T$_n$ indicates the nth triangular number.
I guess the underlying idea I've used here lies in following the natural number sequence across the rows, and down the columns for the digits e and f also.
Best Answer
You have six slots. In each slot you can place any digit between 0 and 9 (ten possibilities). This generates all $10^6$ possible numbers from 0 to $999,999$ (ignoring leading zeros as necessary). To generate a number that doesn't contain a nine, there are only nine possibilities for each slot (namely, $0$ through $8$), yielding $9^6$. So the total number of integers between zero and $999,999$ that do contain the digit nine is $10^6-9^6$. This is the same as the number of positive integers between 0 and a million that do so.