Qua locus

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The problem with dinner parties…

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Mr and Mrs Smith are having a dinner party, and have invited Mr and Mrs Armstrong, Mr and Mrs Brown, Mr and Mrs Cuthbert and Mr and Mrs Dixon. As people arrive, they shake hands, always keeping to the following rules of etiquette:

  • No one shakes their own hand
  • No one shakes their partners hand
  • No pair shakes hands more than once

After the dinner, Mr Smith asked everyone how many times they had shaken hands this evening, and discovered that they had all done it a different number of times.

How many times did Mrs Smith shake hands?

Written by CGF

May 20, 02012 at 19:54

Posted in Puzzles

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8 Responses

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  1. Hint: What if the dinner party had been smaller, and only Mr and Mrs Armstrong had come – but the same rules of etiquette had been observed, and Mr Smith discovered that all the other people (i.e. Mrs Smith, Mr A, and Mrs A) at the dinner had shaken hands a different number of times. How many people did Mrs Smith shake hands with in this instance?

    I love the air kisses and cups of tea and other suggestions from Facebook though 🙂


    May 21, 02012 at 07:12

  2. 8 times?

    Jenny Hare

    May 22, 02012 at 09:17

    • Afraid not…


      May 22, 02012 at 22:25

  3. My guess is 4, but I do not have a proof.


    May 26, 02012 at 09:32

  4. I second 4, but I have proof 😉
    Are we right?

    Alex Robinson

    May 27, 02012 at 11:45

    • But a proof invisible is no proof at all… 🙂

      Christo Fogelberg

      May 27, 02012 at 12:34

      • OK, due to the pigeonhole principle there must be two people who shake hands the same number of times (someone couldn’t of shook hands with everybody and someone else with nobody). Hence Mr Smith must have shook hands the same number of times with someone else.

        Now considering the simple case of only four(two guests) the only way they can shake hands, conforming to the rules, is by Mr Armstrong shaking hands with both Mr and Mrs Smith. Then we have Mr Smith once, Mrs once, Mr Armstrong twice and Mrs Armstrong zero.

        Scale this up to ten people(eight guests) gives and answer of four for Mrs Smith.

        Basically the solution is that when each pair of guest arrive the male shakes hands with everyone(male and female) who is currently in the room. This, in fact, work symmetrically around the male and females so it could be the other way around.


        Alex Robinson

        May 30, 02012 at 09:41

      • Nice… I’ve never thought of the problem in an incremental way like that before, but it works. It generalises nicely to the n-couple version of the problem too.

        I’ve always thought about it like this:

        • By the problem’s description and rules, we know that the other 9 guests shook hands 0, 1, 2, … 8 times.
        • 8 shook hands with everyone they could shake hands with, therefore by a process of elimination 0 and 8 are married
        • The only candidates for 7’s partner are 0 (eliminated already) and 1 because they shook hands with everyone else. We know that the only person 1 shook hands with was 8, therefore 7 and 1 must be married.
        • Carrying this logic forward leaves 4 in the middle as the odd one out. Since Mr Smith is now the only one without a partner, 4 must be Mrs Smith

        Christo Fogelberg

        May 30, 02012 at 21:31

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