Three married couples arrange a party. They arrive at the party one at a time, the couples not necessarily arriving together. They all, upon arriving, shake the hand of everyone already there, except their own spouse. When everyone has arrived, someone asks all the others how many hands they shook upon arriving, and gets five different answers. How many hands did he himself shake upon arriving?
The only possible numbers of hands shaken are 0, 1, 2, 3, 4
since there are only 6 people and no one shakes with self or spouse.
If the asker got 5 different answers, they must be exactly 0, 1, 2, 3, 4
and he must duplicate one of those.
Name the couples A, B, C in order of arrival of their first members.
If both members of a couple arrive among the first three,
then the number of shakes there will be 0, 0, 2 (for AAB)
or 0, 1, 1, (for ABA or ABB).
In that case, both members of couple C arrive among the last three,
and their shakes will be 2, 4, 4 (for xCC)
or 3, 3, 4 (for CxC or CCx). (x here means "A or B").
In any combination of those, there will be two duplications,
which would mean the asker got at least one duplicate response.
Therefore, the first three to arrive must be from separate couples,
(i.e., ABC) and the last three to arrive must be from separate couples (ABC
in any order).
Then, the sequence of shakes is 0, 1, 2, 2, 3, 4
and the asker must be the third or fourth person to arrive,
and therefore he shook 2 hands.
Here are all the possible shake sequences,
and representative arrival sequences for them:
002244 -> AABBCC
002334 -> AABCBC or AABCCB
011244 -> ABBACC or ABABCC
011334 -> ABBCxx or ABACxx
012234 -> ABCxxx
We can also note that total number of handshakes is 12 (6 x 4 / 2)
and since 0 + 1 + 2 + 3 + 4 = 10, there are 2 unaccounted for,
which must be those of the asker.
