What is the minimum number of handshakes at a party with eight people?

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  • Thread starter Ackbach
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    2015
In summary, the minimum number of handshakes at a party with eight people can be calculated using the formula n(n-1)/2, which in this case would be 28. This formula is derived from the fact that each person needs to shake hands with n-1 other people, and is divided by 2 because each handshake is counted twice. The formula only works for even numbers, so for odd numbers, the formula is n(n-1)/2 + (n+1)/2 = (n^2+n)/2. The minimum number of handshakes can be visualized using a complete graph, where 8 people would have 28 handshakes. As the number of people increases, the minimum number of
  • #1
Ackbach
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Here is this week's problem:

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Eight total people (four couples), including the host and hostess, arrive at a party, and a number of handshakes occur. No one shakes hands with their own spouse or with themselves, and everyone shakes at least one hand. The host asks the others how many hands they shook, and everyone (except possibly the hostess) gives a different answer. The hostess is the last to answer; what is the smallest possible number of hands she shook?

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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  • #2
Congratulations to Fallen Angel for a correct solution, which you can read below:

At first, we got 6 different answer (from the couples that aren't host), which is a number from 1 to 6 (bacause no one can shake itself or it's couple and everybody shakes at least once), so all this numbers are someone's answer.

Claim: Hostess just need to shake once.

Proof:

Let $\{m_1,m_2,m_3,f_1,f_2,f_3,\text{host},\text{hostess}\}$ be the vertex of a graph where $m_i-f_i$ is a couple, then we can got the following edges (that means shaking hands).

$\{(\text{host},m_1),(\text{host},m_2),(\text{host},m_3),(\text{host},f_3),(m_1,m_2),(m_1,m_3),(m_1,\text{hostess}),(m_1,f_3),(m_1,f_2),(m_2,m_3),(m_2,f_3),(m_2,f_1),(m_3,f_2)\}$

and we are done because $\deg(m_i)=7-i$, $\deg(f_{i})=i$, $\deg(\text{host})=4$ and $\deg(\text{hostess})=1$.
 

1. How do you calculate the minimum number of handshakes at a party with eight people?

The minimum number of handshakes at a party can be calculated using the formula n(n-1)/2, where n is the number of people. In this case, n=8, so the minimum number of handshakes would be 8(8-1)/2=28.

2. Why is the formula for calculating handshakes n(n-1)/2?

This formula is derived from the fact that in a group of n people, each person needs to shake hands with n-1 other people to have a complete set of handshakes. However, this counts each handshake twice (once for each person involved), so we divide by 2 to get the total number of handshakes.

3. Can the minimum number of handshakes be different if the number of people at the party is odd?

Yes, the minimum number of handshakes will be different if the number of people is odd. This is because the formula n(n-1)/2 only works for even numbers. For odd numbers, the formula is n(n-1)/2 + (n+1)/2 = (n^2+n)/2.

4. Is there a way to visualize the minimum number of handshakes at a party?

Yes, there is a way to visualize the minimum number of handshakes using a graph theory concept called a complete graph. In this case, a complete graph with 8 vertices (representing the 8 people) would have 28 edges (representing the 28 handshakes).

5. How does the minimum number of handshakes change as the number of people at the party increases?

As the number of people at the party increases, the minimum number of handshakes increases as well. This is because there are more possible handshakes between a larger number of people. The relationship between the number of people and the minimum number of handshakes follows a quadratic pattern, increasing by n/2 each time n increases by 1.

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