Bunuel
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 siblings 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?
A. 324
B. 441
C. 630
D. 648
E. 882
Each of the 9 × 2 = 18 members of the twins group shook hands with 16 members of the twins group [the total number of twins – himself – his sibling], so this represents a total of
18 × 16 / 2 = 144 handshakes [We have to divide by 2 because person A’s handshake with person B is the same handshake as person’s B handshake with person A.]
Furthermore, each of the 18 members of the twins group shook hands with 6 × 3 / 2 = 9 members of the triplets group. [Of course, this number also accounts for all the handshakes the members of the triplets group made with the members of the twins group.] So, this is a total of 18 × 9 = 162 handshakes.
Finally, each of the 6 × 3 = 18 members of the triplets group shook hands with 15 members of the triplets group [the total number of triplets – himself – his two siblings], so this represents a total of
18 × 15 / 2 = 135 handshakes [Again, we have to divide by 2 because person A’s handshake with person B is the same handshake as person’s B handshake with person A.]
Therefore, the total number of handshakes is:
144 + 162 + 135 = 441
Answer: B