A bag contains four pieces of paper, each labeled with one of the digits $$1$$, $$2$$, $$3$$ or $$4$$, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $$3$$? ($$2007$$ AMC $$8$$ Problem, Question \#$$24$$) 
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The combinations of digits that give multiples of $$3$$ are $$(1$$, $$2$$, $$3)$$ and $$(2$$, $$3$$, $$4)$$. For each of them, there are $3\times2\times1=6$ possibilities. Thus, there are $6+6=12$ possibilities in total. The number of ways to choose three digits out of four is $$4\times3\times2=24$$. Therefore, the probability is $$\frac{12}{24}=\frac12$$. 