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2011 AMC 12A Problem 9

Problem 9 of 25EasierCombinatoricsProblem-Solving Techniques

At a twins and triplets convention, there were 99 sets of twins and 66 sets of triplets, all from different families. Each twin shook hands with all the twins except his/her sibling and with half the triplets. Each triplet shook hands with all the triplets except his/her siblings and with half the twins. How many handshakes took place?

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Solution

There are 1818 twins and 1818 triplets. Twin-twin handshakes: each twin shakes 18−2=1618 - 2 = 16 other twins, giving 18⋅162=144.\dfrac{18 \cdot 16}{2} = 144. Triplet-triplet handshakes: each triplet shakes 18−3=1518 - 3 = 15 other triplets, giving 18⋅152=135.\dfrac{18 \cdot 15}{2} = 135. Twin-triplet handshakes: each twin shakes half the 1818 triplets, giving 18⋅9=16218 \cdot 9 = 162 (each such handshake counted once). The total is 144+135+162=441.144 + 135 + 162 = 441. Thus, the correct answer is B.
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