2011 AMC 10A Problem 21

There are two cases: either both selected pairs contain only genuine coins or each selected pair has one counterfeit coin. For the first case, there are $\binom{8}{2} = 28$ ways to choose the coins for the first pair and $\binom{6}{2} = 15$ choices for the second pair. We also have to divide by $2$ since we can swap the pairs. This gives us $$28 \cdot 15 \div 2 = 210$$ configurations for this case. For the second case, there are $\binom{8}{2} = 28$ ways to choose the non-counterfeit coins. There is only one choice for the counterfeit coins. There are two ways to create the two pairs, two choices for which counterfeit coin goes with a genuine coin. This means that there are $28 \cdot 2 = 56$ configurations for this case. The desired probability is then $$\dfrac{210}{210 + 56} = \dfrac{210}{266} = \dfrac{15}{19}.$$ Thus, D is the correct answer.