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John and Mary go to Dinner

  1. Sep 18, 2007 #1
    [SOLVED] John and Mary go to Dinner

    Hi, I have this homework problem and so am posting it alongwith my suggested solution:


    Problem Statement:

    John and Mary are Married. They agree to go to dinner with four other married couples. so there are 5 couples, or 10 people altogether. They arrive at the restaurant, and because there are some people who do not know other people, there are some handshakes that take place and some introductions. Anyone who does not know anybody will shake hands with that person. People will not shake hands with someone they already know and ofcourse, you do not shake hands with your own spouse. At the end of the dinner, John asks all the people present, including his own wife, how many hands each of them shook. He gets the following answers 0,1,2,3,4,5,6,7 and 8. There are 9 other people there, and they give him 9 different answers. How many hands did John's wife Mary shake? Prove your answer


    My attempted solution:

    Out of a total of 10 people, Johns wife would not shake hands with herself and her spouse, in this case John. Hence she would have shook hands with 10 -2 = 8 people.
     
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  3. Sep 18, 2007 #2

    EnumaElish

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    Maybe I am slow, but what if Mary knew everybody and so did not shake any hands? How do you exclude this possibility?
     
  4. Sep 18, 2007 #3
    You are right, that did not come to me. However looking at the problem more closely, there is a person who shook hands 0 times, meaning that person knew everybody there. That would imply to be Mary, but how would I be able to prove that answer. If we take into consideration the person who shook hands 0 times that means the person knew all the other couples there (excluding himself/herself and spouse), changing the answer to 10 - 3 = 7.
     
    Last edited: Sep 18, 2007
  5. Sep 18, 2007 #4

    learningphysics

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    Hint: Find out who's married to who... The person who shook hands with 8 people... who didn't he shake hands with?
     
  6. Sep 18, 2007 #5
    Would this assumption be incorrect that the person who shook hands with 8 people would not have shook hands with the spouse and himself/herself?
     
  7. Sep 18, 2007 #6

    learningphysics

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    We are already certain this is true. We know that nobody shakes hands with themselves or their spouses.
     
  8. Sep 18, 2007 #7
    Based on the problem, how would I be able to find out who is married to who...I would just need a little direction, it would not be possible to construct a truth table as that would not be the right approach.
     
  9. Sep 18, 2007 #8

    learningphysics

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    Who is the person who shakes 8 people's hands married to? This is the first step. Hint: Who doesn't he shake hands with?
     
  10. Sep 18, 2007 #9
    Upon reading the question and logically pondering, I think that Mary shook 0 hands based on:

    They had initially agreed to go to dinner with 4 other couples, after the dinner it is John who takes the lead in asking questions, which implies that John knows everybody which further implies that Mary knows the guests, hence she shook 0 hands.
     
  11. Sep 18, 2007 #10

    EnumaElish

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    I am not sure that your question has enough information. Why is John & Mary is not perfectly substitutable by Henry & Alice?
     
  12. Sep 18, 2007 #11
    While that substitution may be valid, I base my assumptions on the fact that John was the one to ask the question, which implies that he knows all the guests so we can also put this in the form of the following:

    p=John
    q=Mary

    using p->q then if p then q hence true, so if John knows all the guest then so does Mary.

    This is my assumption. I myself am looking for the answer, as for the question, this is what has been given to us by our Professor as a bonus point.
     
  13. Sep 18, 2007 #12

    learningphysics

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    No, that isn't the answer.

    Which person does the person who shook 8 hands married to? Which person didn't he shake hands with? There's only two people out of 10 that he didn't shake hands with... himself and his wife.
     
  14. Sep 18, 2007 #13
    If one person shook 0 hands, that means his/her spouse shook 8 hands because the spouse would be the only person with 8 options open.

    Couple A: 0/8

    And since John asked the question and got those 9 answers, that disqualifies the 0/8 couple.

    So Mary shook between 1-7 hands.

    The person who shook 1 hand shook the person who shook 8 hands. As did everybody else except for the 8 shaker's spouse.

    And those are the only answers that can be derived(at least I think) from the information given. I don't see any other limitations that can be placed.

    Just because John spoke at the end doesn't mean he knows everybody. In a real life situation, it is very possible for a completely new guy to ask that question.
     
  15. Sep 18, 2007 #14

    learningphysics

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    You can keep going. Who is the 7 shaker married to?
     
  16. Sep 18, 2007 #15
    If we go by the above assumption, for the person who shook 2 hands shook everybody's hand including shaker 8 and 7. That still would not be accurate.

    Could somebody please help me in getting a more precise method of determining the solution?
     
  17. Sep 18, 2007 #16

    learningphysics

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    This is how I did it.

    The person who shook hands with 8 people didn't shake hands with 2 people. One is himself. The other is his wife.

    The 0 shaker didn't shake hands with the 8 shaker. Therefore the 0 shaker must be the 8 shaker's husband/wife.

    The 1 shaker only shook hands with the 8 shaker. He didn't shake hands with the 7 shaker.

    The 7 shaker shook hands with everyone, except himself, the 0 shaker and his wife. Therefore the 1 shaker must be the 7 shaker's wife/husband.

    The 2 shaker only shook hands with the 8 shaker and 7 shaker. He didn't shake hands with the 6 shaker

    The 6 shaker shook hands with everyone, except himself, the 0 shaker, the 1 shaker and his wife. Therefore the 2 shaker must be his wife.

    etc....
     
  18. Sep 18, 2007 #17
    Hi,

    Taking up from where you left off, I only rearranged the lines for my understanding and have come up with the following:

    The 0 shaker didn't shake hands with the 8 shaker. Therefore the 0 shaker must be the 8 shaker's husband/wife.

    The 8 shaker shook hands with everyone except himself and his wife.

    The 1 shaker only shook hands with the 8 shaker. He didn't shake hands with the 7 shaker.

    The 7 shaker shook hands with everyone, except himself, the 0 shaker and his wife. Therefore the 1 shaker must be the 7 shaker's wife/husband.

    The 2 shaker only shook hands with the 8 shaker and 7 shaker. He didn't shake hands with the 6 shaker

    The 6 shaker shook hands with everyone, except himself, the 0 shaker, the 1 shaker and his wife. Therefore the 2 shaker must be his wife.

    The 3 shaker shook hands with 8,7 and 6. He didnt shake hands with shaker 5

    The 5 shaker shook hands with everyone, except himself, 0,1,2 and his wife, therefore shaker 3 must be his wife


    That leaves shaker 4, who must be Mary, John's wife and shook hands with exactly four people.


    Am I right?
     
  19. Sep 19, 2007 #18

    learningphysics

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    Yes, exactly.
     
  20. Sep 19, 2007 #19

    EnumaElish

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    And just how do you know the 4-shaker is John's wife Mary, and not Henry's wife Alice? That John & Mary are not the 5-shaker and the 3-shaker, or vice versa?
     
    Last edited: Sep 19, 2007
  21. Sep 19, 2007 #20

    Taking a look at the problem from the beginning....there is a sum of 8 in every combination...0/8, 1/7,2/6,3/5, if we go with this pattern and get to the middle i.e. 4 there is no one else to make a pair with, all the pairs have been done, the remaining shaker 4 then would be paired with someone out of the pair, in this case, 9. So, if John is 9 then Mary is 4. Counting ahead from 4 to 8 (leaving 9 out because that is her spouse and herself which makes 10), we get the number 4.
     
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