Question

1 Shaky Party A host and a hostess invited 4 couples to a party and at the end of the party the host asked everyone including his spouse how many hands they shook during the introduction period of the party. Everyone gave a different answer ranging from 0 handshakes up to 8 .

Although we have no idea who shook hands with whom , we know know that no one shook hands with themselves and no one shook hands with his or her own spouse. Given those fact, determine how many hands did the hostess shook?

Answer 1

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

So, 4 is the answer.