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Math Help - Confused on what to do here..

  1. #1
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    Confused on what to do here..

    Five friends met for lunch, and they all shook hands. Each person shook the other persons right hand only once. What was the total number of handshakes?

    Would you add 5+5?
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  2. #2
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    Quote Originally Posted by eh501 View Post
    Five friends met for lunch, and they all shook hands. Each person shook the other persons right hand only once. What was the total number of handshakes?

    Would you add 5+5?
    No. Let A B C D E be the people. Then A shakes hands with B,C,D,E. Then B shakes hands with C,D,E but not A because they already shook hands. Then C shakes hands with D,E. And D shakes hands with only E.
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    How would you know when to do that, when to multiply or when to add?
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    Quote Originally Posted by eh501 View Post
    How would you know when to do that, when to multiply or when to add?
    Each problem is different, you need to understand how to approach a problem and decide whether you have to add or multiply.

    You can also do it another way. Draw a picture and the lines represent the people who shook hands. Then count the number of lines.
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  5. #5
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    Ok, so whats the answer?
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    Quote Originally Posted by eh501 View Post
    Ok, so whats the answer?
    You are going to find the answer. I basically gave you the solution.
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    I know you did but i really don't know what to do..do i just add up the letters? I still came up with 10.
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    Quote Originally Posted by eh501 View Post
    I know you did but i really don't know what to do..do i just add up the letters? I still came up with 10.
    Yes, 10 is the right answer. If you add up the lines. But your reasoning that 5+5=10 is wrong. Try for example 4 people instead of 5. Then the answer would be 3+2+1 = 6 and not 4 + 4 = 8.
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  9. #9
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    If n people shake hands with each other such each pair of persons shake hands once and only once, the total number of handshakes is ^n\mathrm{C}_2=\frac{n(n-1)}{2}.
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