# Puzzle | Handshakes

A couple invites n – 1 other couples to dinner. Once everyone arrives, each person shakes hands with everyone he doesn’t know. Then, the host asks everyone how many hands they shook and each person replies with a different number. Assuming that everyone knows his or her own spouse, how many hands did the hostess shake?

**Solution:**

N – 1, where N is the total no. of couples at the dinner.**Explanation:**

- The possible numbers of handshakes range from 0 to 2N-2. (2N-1 would require that a person shook hands with every other person at the party, but nobody shook hands with his/her spouse.)

There are 2N-1 different numbers, and the host got 2N-1 different answers, so every number is represented. - One person (0) shook no hands, and another (2N-2) shook hands with everybody from all the other couples. This is only possible if these two are a married couple, because otherwise 2N-2 would have had to have shaken 0’s hand.
- One person (1) shook only 2N-2’s hand, and another (2N-3) shook hands with everybody from all the other couples except 0. Again, these two must be married, or else 2N-3 would have had to have shaken 1’s hand, a contradiction.
- Continuing this logic, eventually you pair up all the couples besides the hosts, each one pairing a shook-no-hands-not-already-mentioned person with a shook-all-hands-not-already-mentioned person, the last having shaken N-2 and N hands respectively.
- This tells that the hostess must have shaken N-1 hands, since there are N-1 shook-no-other-hands people and N-1 shook-all-other-hands people. Both the host and hostess shook hands with exactly one member of each couple – the same ones – and thus each shook N-1 people’s hands

**Example:**

N = 6

- 5 couples are invited to the dinner. Among the total six couples, no one shook more than 10 hands.
- Therefore, if eleven people each shake a different number of hands, the numbers must be 0, 1, 2, …, and 10.
- The person who shook 10 hands has to be married to the person who shook 0 hands (otherwise that person could have shaken only ten hands).
- Similarly, the person who shook nine hands is bound to be married to the person who shook 1 hand.
- Continuing the logic, couples shook hands in pairs as mentioned 10/0, 9/1, 8/2, 7/3, 6/4. The only person left who shook hands with 5 is the hostess.