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 are actually counting the number of subsets of $\{A,B,C\}$ of a fixed size ($2$, in your example), without regarding order, and allowing duplicates. A more general way to look at it, is taking some set $A=\{1,\ldots,n\}$, and trying to assign a number of occurrences to each of $1\ldots n$, that is, "$1$ appears once, $2$ appears five times, etc.", while demanding that the sum of these numbers be some constant $k$ - that is, you're only selecting a set of $k$ elements.
This is equivalent to the following situation:
You have $n$ buckets and $k$ balls. You wish to toss the balls into the buckets, putting as many balls in one bucket as you like. In order to calculate the number of ways to do that, it's more convenient to look at the $n$ buckets as $n-1$ partitions - defining $n$ different spaces, that you can place among the balls, like in the following illustration:
$$\cdot|\cdot\cdot\cdot||\cdot|\cdot\cdot||$$
This has $7$ buckets, with $1,3,0,1,2,0,0$ balls respectively.
Now we can give a general equation.
Sort the partitions and the balls together - there are $(n-1+k)!$ ways to do that.
Disregard the internal ordering of the partitions and the internal ordering of the balls, as you're only interested in the number of balls in each "bucket", and the "buckets" are defined by their position regardless of which partitions surround them.
Therefore, the number of ways to select a set of size $k$, with duplicates allowed and without regarding order, from a set of size $n$ is $\frac{(n+k-1)!}{(n-1)!k!}=\binom{n+k-1}{k}$.
In your particular case, $n=3$ and $k=2$, so this equals $\binom{4}{2}=6$.