Answer: Choice C) 220
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We have 12 choices for the first selection, 11 for the second, and 10 for the third. There are 12*11*10 = 1320 permutations. If your teacher was asking about permutations, then you would be done at this point. However, your teacher is asking about combinations. With combinations, order does not matter.
For any group of 3 items, there are 3! = 3*2*1 = 6 ways to arrange this group. This means that we must divide 1320 over 6 to correct for the fact that we overcounted by a factor of 6
In this case,
number of combinations = (number of permutations)/6
number of combinations = 1320/6
number of combinations = 220
More generally, I'm using the connection that
nCr = (nPr)/(r!)