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Homework Help: Combination question

  1. Jan 2, 2008 #1
    1. The problem statement, all variables and given/known data

    There're 8 people in the room and each person shakes hand with the other .

    2. Relevant equations

    How many hand shakes are there ?

    3. The attempt at a solution

    One person will shake hand with 7 other people and so on . As a result there should be 8 power of 7 but the result is different ?

    P/S : Can you explain what's the difference between combination and permutation ?
  2. jcsd
  3. Jan 2, 2008 #2
    First of all take the easy way!

    Let's say that there are 3 people. How many hand shakes actually are, and how many do you calculate with your method?
  4. Jan 2, 2008 #3
    Supposing there're 3 people . One person shakes hand with 2 others ? and there're 9 hand shakes using my method and is it actually 2 hand shakes ? What's the equation for this ?
  5. Jan 2, 2008 #4


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    Draw a ring or circle and start arranging equally located points, starting with 3 points on the circle. Now, draw all connecting segments between all points. Count the segments. Continue doing this using 4 points, then 5 points, up until 8 points. You should determine what the resulting number pattern is, and from this, you can determine how many segments will connect any number of equally distributed points around the circle.

    This is also known as "the handshake problem", and it has been discussed in the forums previously.
  6. Jan 3, 2008 #5
    It is actually 3 hand shakes. Call the people A, B, C (stupid names! :smile:)

    Man A, shakes B and C (2 hand shakes)

    Now we don't have to bother with A any more, he/she did his/her job

    We are left with B and C (1 hand shake) Total = 3
    Ok, up to this?
  7. Jan 3, 2008 #6


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    Does "8 power of 7" mean 87? And how did you arrive at that?

    Each person shakes hands with every other person. If there are n people, each person must shake hands with n-1 other people and so there are n(n-1) handshakes. EXCEPT each handshake involves two people so that "overcounts".
  8. Jan 3, 2008 #7
    I actually got it myself . One person shakes hand with 7 others . And there're 8 times of that action ? ( Because we have 8 people ) so 42 ? . But i found it nothing to do with combination nor permutation . What's the difference between them ?

    By the way does anybody know what is alpha-numeric plate ?
  9. Jan 3, 2008 #8


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    No, you haven't got it. First of all 8(7) is NOT 42, it is 56! If 8 people each shake hands with 7 people that would be 56 handshakes except that two people participate in each handshake- you are counting each handshake twice.

    Another way to do it, but more tedious, would be to argue that person "A" shakes hands with each of the other people: 7 handshakes. Person "B" then shakes hands with 6 other people (his handshake with "A" already being counted), then person "C" shakes hands with 6 new people, etc. The number of handshakes is 7+ 6+ 5+ 4+ 3+ 2+ 1.
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