Alice and Bill threw a party and invited four other couples. As each couple arrived there were greetings, including handshakes.
Later in the evening, Bill asked everyone, including Alice how many people they shook hand with. Every answer was different. No one shook hands with his or her own partner.
How many hands did Alice shake?
(In reply to
Puzzle Answer by K Sengupta)
Since there were 10 people and no one shook hands with his/her spouse, it follows that the maximum number of handshakes a person could have made is 8 and the corresponding minimum is 0.
Let us denote:
P(n) = the person having shaken hands with n persons, when n= 0 to 8 inclusively.
Arranging this in order, we have:
P(0), P(1), P(2), P(3), P(4), P(5), P(6), P(7), and P(8)
So, P(8) shook hands with everybody except his/her spouse, and accordingly P(0) must be P(8)'a spouse.
Now, obviating the couple {P(0), P(8)} out of our consideration, we follow the same logic and discern that the pair {P(1), P(7)} must correspond to another married couple. Proceeding similarly, we observe that:
{P(2}, P(6)} and {P(3), P(5)} are two other married couples.
Hence, P(4)'s spouse also shook hands with 4 person, and represent a second P(4). Therefore {P(4), P(4)} also represents a married couple.
Now, we observe that Bill was the questioner and did not respond to his own query.Therefore, he must have shaken hands with precisely 4 persons, and therefore represents the second P(4). The first P(4) must therefore be his wife Alice.
Consequently Alice shook precisely 4 hands.
Edited on September 12, 2022, 11:41 pm
Explanation to Puzzle Answer
