Question

At a twins and triplets convention, there were $9$ sets of twins and $6$ 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?

$324$

$441$

$630$

$648$

$882$

Solution:

There are $18$ twins and $18$ triplets.

Twin-twin handshakes: each twin shakes $18 - 2 = 16$ other twins, giving $\dfrac{18 \cdot 16}{2} = 144.$

Triplet-triplet handshakes: each triplet shakes $18 - 3 = 15$ other triplets, giving $\dfrac{18 \cdot 15}{2} = 135.$

Twin-triplet handshakes: each twin shakes half the $18$ triplets, giving $18 \cdot 9 = 162$ (each such handshake counted once).

The total is $144 + 135 + 162 = 441.$

Thus, the correct answer is B.

2011 AMC 12A Problem 9

Solution:

Problem 9 in Other Years