Hi, demeraldaga:
The smallest possible designs that you're asking about (each pair of
points appears but once) are called Steiner triple systems. Here's a
link which gives the basic result about when they exist:
[Steiner Triple System  from MathWorld]
http://mathworld.wolfram.com/SteinerTripleSystem.html
These are special cases of balanced incomplete block designs, namely
those for which the block size is 3 and the number of times each
distinct pair appears is 1. There are a number of notations used,
but it will be convenient to refer to these as (v,3,1)designs. The
general notation is:
(v,k,?)designs
where v = number of points (or lists in your application),
k = size of each block (which is 3 in your application),
? = number of times each disinct pair appears (1 is minimal).
The theorem proved by T. P. Kirkman in 1847 is that (v,3,1)designs
exist exactly when the remainder of v divided by 6 is either 1 or 3:
v ? 1,3 mod 6
So the most minimal designs exist only for 7 (which you know about)
and 9 (which we'll show in a moment; it's given in the link above),
and then for every multiple of 6 added to either 7 or 9.
v = 7,9,13,15,19,21,25,...
A solution for 9 points or lists is unique (up to rearranging the
points) and where the solution you found is called the Fano plane,
this is called an affine plane:
{{a,b,c},{d,e,f},{g,h,i},
{a,d,g},{b,e,h},{c,f,i},
{a,e,i},{b,f,g},{c,d,h},
{a,f,h},{b,d,i},{c,e,g}}
This will seem less mysterious if you compare the triples shown to the
rows, columns, and (wraparound) diagonals of this 3x3 array:
a b c
d e f
g h i
Thus exactly one "line" through every pair of (distinct) points.
While the solutions for v = 7,9 are essentially unique (up to
"isomorphism"), the number of nonisomorphic solutions grows rapidly.
For v = 13 there are just two different solutions, then for v = 15
there are 80!
Here's one way to construct a solution for v = 13. Consider these two blocks:
{0,1,4} and {0,2,8}
Now for each nonzero "residue" modulo 13, we add (mod 13) to each
entry in both blocks to get two more blocks. The result is a
collection of 26 "triples" that form a (13,3,1)design:
{{0,1,4},{0,2,8},{1,2,5},{1,3,9},{2,3,6},{2,4,10},
{3,4,7},{3,5,11},{4,5,8},{4,6,12},{5,6,9},{5,7,0},
{6,7,10},{6,8,1},{7,8,11},{7,9,2},{8,9,12},{8,10,3},
{9,10,0},{9,11,4},{10,11,1},{10,12,5},{11,12,2},{11,0,6},
{12,0,3},{12,1,7}}
This construction is an example of using the method of differences.
What makes it tick is that every possible nonzero difference 112 is
produced when two values from the same one of our two "initial" blocks
are subtracted mod 13. By virtue of this every pair of points appears
in some block.
Here's a quick sketch of a way to construct a (15,3,1)design. Take
the nonzero binary 4bit words as your points:
{ 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000,
1001, 1010, 1011, 1100, 1101, 1110, 1111 }
and define a triple of this points {X,Y,Z} whenever:
X xor Y = Z
It's a property of the "exclusive or" operation that the above is
equivalent to saying either X xor Z = Y or Y xor Z = X, so that these
triples form welldefined subsets of size 3 (blocks).
Now let's back up and consider what our options are for other values:
v = 6, 8, 10, 11, 12
For some general results on balanced incomplete block designs, see the
introductory sections of:
[On the Existence of (v, 5, 1)BIBDs by Clayton Smith]
http://www.argilo.net/papers/bibd.pdf
In particular Lemma 3 tells us that each point in a (v,k,?)design
will appear exactly:
(v  1)
r = ? 
(k  1)
times, ie. a whole number of times (the same number for each point),
and will require this many blocks (the number of points, times how
many appearance each makes, divide by the size of a block):
b = vr/k
Now for v = 6 and k = 3, we need ? even to get a whole number. The
smallest possibility is ? = 2, so that each point appears r = 5 times.
The number of blocks is then b = 6*5/3 = 10.
One such (6,3,2)design is this collection of ten blocks:
{{a,b,c},{a,c,d},{a,d,e},{a,e,f},{a,b,f},
{b,d,e},{b,c,e},{b,f,d},{c,d,f},{c,e,f}}
in which you can check each symbols appears five times and each
distinct pair twice.
That's still better than the complete block design on v = 6 points,
which requires 20 blocks.
However for v = 8 we cannot do better (with k = 3) than taking the
complete block design. To see why, first look at the relationship on
number of times r that each point appears:
(v  1)
r = ?  = ? * (7/2)
(k  1)
so again ? will need to be even. But also count the number of blocks:
b = vr/k = r * (8/3)
which implies that r must be divisible by 3. That requires (from
above) that ? is divisible by 3 as well as 2, i.e. ? must be at least
6. And ? = 6 gives us just the 56 blocks of the complete block
design.
For v = 10 we already know that we need at least ? = 2. The following
example shows this is possible, albeit using a bit different technique
than what you might have expected!
So far our examples have not used the same _triple_ more than once.
In other words all our blocks were different. This next example of a
(10,3,2)design contains some _repeated_ blocks. However every
distinct pair appears just twice among all the 30 blocks:
{{1,2,3},{1,2,4},{1,3,4},{2,3,4},
{1,5,6},{1,5,7},{1,6,7},{5,6,7},
{1,8,9},{1,8,10},{1,9,10},{8,9,10},
{2,5,8},{2,5,8},{2,6,9},{2,6,9},
{2,7,10},{2,7,10},{3,5,10},{3,5,10},
{3,6,8},{3,6,8},{3,7,9},{3,7,9},
{4,5,9},{4,5,9},{4,6,10},{4,6,10},
{4,7,8},{4,7,8}}
Despite the repeated blocks this is considerable better than the
complete block design, which for v = 10 requires 120 blocks. For the
construction of this particular example, see here:
[Pairwise balanced designs]
http://www.math.mun.ca/~davidm/4341/pbd.pdf
The analysis we did before by counting how many times r that each
point appears and how many blocks b there are (which must both be
whole numbers) tells us this about the remaining cases v = 11,12:
If v = 11, then ? must be at least 3.
If v = 12, then ? must be at least 2.
I'll need to do some further research to find designs of this types however.
regards, mathtalkga 
Clarification of Answer by
mathtalkga
on
01 Oct 2004 07:55 PDT
Hi, demeraldsga:
Here are a few more designs in which you may have an interest. I've
labelled the points beginning with 0. These are among a large number
of 2designs cataloged at this Web site:
[Database at DesignTheory.org]
http://www.designtheory.org/database/
The first design has triples from 10 points in which every pair
appears twice (and each point is used 9 times), but unlike my earlier
example, this one has no repeated "blocks" (all the blocks are
different):
{ 0, 1, 2 } { 0, 1, 3 } { 0, 2, 3 } { 0, 4, 5 } { 0, 4, 6 }
{ 0, 5, 7 } { 0, 6, 8 } { 0, 7, 9 } { 0, 8, 9 } { 1, 2, 3 }
{ 1, 4, 5 } { 1, 4, 9 } { 1, 5, 8 } { 1, 6, 7 } { 1, 6, 8 }
{ 1, 7, 9 } { 2, 4, 6 } { 2, 4, 9 } { 2, 5, 7 } { 2, 5, 8 }
{ 2, 6, 7 } { 2, 8, 9 } { 3, 4, 7 } { 3, 4, 8 } { 3, 5, 6 }
{ 3, 5, 9 } { 3, 6, 9 } { 3, 7, 8 } { 4, 7, 8 } { 5, 6, 9 }
Here's a design on 11 points in which every pair appears 3 times among
the "blocks" of three (triples), and no repeats among the blocks:
{ 0, 1, 3 } { 0, 1, 4 } { 0, 1, 7 } { 0, 2, 5 } { 0, 2, 7 }
{ 0, 2,10 } { 0, 3, 9 } { 0, 3,10 } { 0, 4, 5 } { 0, 4, 6 }
{ 0, 5, 9 } { 0, 6, 8 } { 0, 6,10 } { 0, 7, 8 } { 0, 8, 9 }
{ 1, 2, 4 } { 1, 2, 5 } { 1, 2, 8 } { 1, 3, 6 } { 1, 3, 8 }
{ 1, 4,10 } { 1, 5, 6 } { 1, 5, 7 } { 1, 6,10 } { 1, 7, 9 }
{ 1, 8, 9 } { 1, 9,10 } { 2, 3, 5 } { 2, 3, 6 } { 2, 3, 9 }
{ 2, 4, 7 } { 2, 4, 9 } { 2, 6, 7 } { 2, 6, 8 } { 2, 8,10 }
{ 2, 9,10 } { 3, 4, 6 } { 3, 4, 7 } { 3, 4,10 } { 3, 5, 8 }
{ 3, 5,10 } { 3, 7, 8 } { 3, 7, 9 } { 4, 5, 7 } { 4, 5, 8 }
{ 4, 6, 9 } { 4, 8, 9 } { 4, 8,10 } { 5, 6, 8 } { 5, 6, 9 }
{ 5, 7,10 } { 5, 9,10 } { 6, 7, 9 } { 6, 7,10 } { 7, 8,10 }
Finally here is a design on 12 blocks in which every pair appears
twice, but this design does have repeated blocks:
{ 0, 1, 2 } { 0, 1,11 } { 0, 2, 7 } { 0, 3, 6 } { 0, 3, 9 }
{ 0, 4, 8 } { 0, 4, 8 } { 0, 5, 7 } { 0, 5,10 } { 0, 6, 9 }
{ 0,10,11 } { 1, 2, 3 } { 1, 3, 8 } { 1, 4, 7 } { 1, 4,10 }
{ 1, 5, 9 } { 1, 5, 9 } { 1, 6, 8 } { 1, 6,11 } { 1, 7,10 }
{ 2, 3, 4 } { 2, 4, 9 } { 2, 5, 8 } { 2, 5,11 } { 2, 6,10 }
{ 2, 6,10 } { 2, 7, 9 } { 2, 8,11 } { 3, 4, 5 } { 3, 5,10 }
{ 3, 6, 9 } { 3, 7,11 } { 3, 7,11 } { 3, 8,10 } { 4, 5, 6 }
{ 4, 6,11 } { 4, 7,10 } { 4, 9,11 } { 5, 6, 7 } { 5, 8,11 }
{ 6, 7, 8 } { 7, 8, 9 } { 8, 9,10 } { 9,10,11 }
regards, mathtalkga
