Hi Neil:

Your request is a little ambiguous.

If you want to know the number of ways in which you can arrange 10 things (in your case, pairs of numbers) three at a time without respect to order, then it is referred to as a "combination and the formula is:
nCr = n!(n-r)!/r!
You are arranging 10 pairs, three at a time so this becomes:
10C3 = 10!(10-3)!/3! which is 120

If you are looking for the smallest number of triples so that each of the pairs gets a look in at least once, then 10/3 tells you it will take 4 triples

Is this more (or less) than you were looking for?

Godfrey Nicholson