A party is held at the house of the Mehtas. There are five other couples present (besides Mr and Mrs Mehta), and many, but not all, pairs of people shook hands. Nobody shook hands with anyone twice, and nobody shook hands with his/her spouse. Both the host and the hostess shook some hands. At the end of the party, Mr Mehta polls each person (other than himself) shook. Each person gives a different answer. Determine how many hands Mrs Mehta must have shaken. (Can we prove that it was not Mrs Mehta who shook 10 hands?)
Answers
Answer:
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Explanation:
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Answer:
Mrs. mehta must have shaken 5 hands !
Explanation:
Deduction 1 From the condition that nobody shook hands with his/her spouse, it is clear that none of the twelve people in the party shook more than 10 hands.
(Since, nobody shakes hands with himself or his/her spouse, it leaves a maximum of 10 people to shake hands with).
Deduction 2 Mr Mehta has asked the question to eleven different people and each of them has given a different answer. Also, the highest answer anyone could have possibly given is 10. Hence, the only way to distribute different numbers of hand shakes amongst the 11 people is:
0,1,2,3,4,5,6,7,8,9,10. [ Note: somebody shook 0 hands and somebody shook 10]. Deduction 3 Since, the host & hostess have both shaken some hands, the person who
shook ‘0’ hands cannot be either M or M. It has to be one of the other 10 people in the party.
[At this point you need to realise that in the context of this problem A,A B,B,C, C, D, D, E & E are alike, i e. there is no logical difference amongst these 10 and you have exactly the same information about each of these 10 people. However, Mrs Mehta is different because she stands out as the hostess as well as the wife of the person who has asked the question].
Since all ten guests are the same, assume ‘A’ shook no hands. This leads us to the following deduction.
Deduction 4 Take any one person apart from A & A; say B. B will not shake hands with himself & his wife. Besides B will also not shake hands with A (who has shaken no hands). Thus, B can shake a maximum of 9 hands and will thus not be the person to shake 10 hands.
What applies to B, applies to B, C, C, D, D, E, E and M.
Hence, A is the only person who could have shaken 10 hands. Hence, amongst the couple A & A, if we suppose that A had shaken 0 hands, then A must have shaken 10 hands.
Note: The main result here is that, out of the people to whom M has asked the question, and amongst whom we have to distribute the numbers 0 to 10 there has to be a couple who has had 0 & 10 handshakes. It could be any of the five couples, but it cannot be M who has either 0 or 10 handshakes.
We now proceed, using the same line of reason as follows.
Deduction 5 Suppose B has 1 handshake — he must have shaken hands with A (who has shaken everybody’s hands she can).
Then, B wouldn’t have shaken hands with anyone out of A, B, C, C, D, D, E, E & M. At this point the following picture emerges:
A— A—10 B—1B— C— C— D— D— EE
M
Numbers left to be allocated — 2, 3, 4, 5, 6, 7, 8, 9.
Considering C, as a general case, he cannot shake hands with C, A (Who shook no hands) & B (Who shook hands only with A. This is mandatory since A has shaken hands with 10 people).
Thus, C can shake hands with a maximum of 8 people and this deduction will be true for C, D, D ,E , E & M too. Hence, the only person who could get 9 handshakes is B.
Thus, we conclude that just like 0 and 10 handshakes were in one pair, similarly 1 and 9 handshakes too have to be part of one pair of husband and wife.
Similar deductions, will lead to the realisation that 2 & 8, 3 & 7 and 4 & 6 handshakes will also occur for couples amongst the 11 people questioned.
Hence, M must have shaken 5 hands.