Solving John and Mary's Handshake Problem

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SUMMARY

The discussion centers on the Handshake Problem involving John and Mary at a dinner party with four other couples. Each individual shook hands with those they did not know, resulting in unique handshake counts from 0 to 8. The conclusion drawn is that Mary shook hands with 4 people, as each number from 0 to 8 corresponds to different individuals, and the logic extends to n couples where the answer is n-1 for the number of handshakes.

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andytran
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hi,

i have this problem for my discreet math class, after an hr and still didn't figure it out so the next best thing is to post it here to get some advices that might jumpstart my brian.
anyway here it is (a bit long).



John and his wife, Mary, went to a dinner party with four other married couples. Because everyone didn't know everyone else, there were the usual introductions and handshakes when the five couples gathered at the table. Each person shook hands with everyone they did not already know, but did not shake hands with anyone they already knew, including their own spouse, of course.

At the end of the dinner, John asked everyone, "How many different people did you shake hands with?" All answered truthfully and each gave him a different numbers: 0,1,2,3,4,5,6,7, and 8.

Determine Mary's answer and give a convincing argument that your answer is correct. Explain your reasoning as clearly and succinctly as you can. It is not neccesary to use formal mathematical or logical notation.

What is Mary's answer when, instead of five couples, there are n couples, in total?


thx
 
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quark said:
I did it in Brain Teaser section. Follow the link.
https://www.physicsforums.com/showthread.php?t=73520
When there are n couples and the handshakes are from 0 to 2n-2 then the answer would be n-1 (following the same logic)

Regards,


wow thx! :)
 

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