I need to find the number of unique combinations in the set 1A, 1B, 1C, 2A, 2B, 2C, 3A, 3B, 3C. Each integer must appear a maximum of once in each combination – i.e. 1A, 1B is invalid. Combinations can have one, two or three values. Order not important.
by 
mpr1977
ANSWERS: 1
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Lets call each alpha-numeric (1a or 2b etc) combo a digit. One digit unique combos is nine, one for each available digit. Two digit unique combo work by taking one digit, lets say 1A then counting the remaining available digits For the ones it is the six remaining digits starting with 2 and 3. So thats eighteen total, three 1's times six. now the 2s and 3s need to be accounted for. For the 2s there are 3 digits remaining for each, 6 total. And those eliminate the 3's as well. So 6 plus eighteen means 24 possible 2 digit combos. For three digit sets you need a single 1,2 and 3 so we only need to account for the ones b/c all other combos are invalid. so 1a, 2a, 3x has 3 possibles. This means that 4each set starting with 1a has 9 possible and that there are 9 combos for each 1s digit. 3x9 is 27. Add the totals for one digit 2 digit and 3 digit sets together. 9+24+27=60 total unique sets. Hopefully you can follow it.
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