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Thread: Hand shake????

  1. #1
    Junior Member abhforum's Avatar
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    Hand shake????

    At a party, everyone shook hands with everybody else.
    There were 66 handshakes. How many people were at the
    party?
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  2. #2
    MHF Contributor

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    Re: Hand shake????

    Quote Originally Posted by abhforum View Post
    At a party, everyone shook hands with everybody else.
    There were 66 handshakes. How many people were at the
    party?
    Can you solve $\dbinom{n}{2}=66~?$
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  3. #3
    Junior Member abhforum's Avatar
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    Re: Hand shake????

    Thanks a lot
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  4. #4
    Junior Member mathmonster's Avatar
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    Re: Hand shake????

    basic drawn out method

    Triangle numbers . . .

    1 person shake hands 0 times (set n=0 as 1 person cannot create a handshake)
    2 people shake hands 1 time set n=1
    3 people shake hands 3 times set n=2
    4 people shake hands 6 times set n=3 etc

    n(n+1) = 66
    2

    n(n+1) = 132

    therefore you want two numbers with a difference of 1 that multiply together to give you 132.

    11x12=132 therefore n=11
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  5. #5
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    Re: Hand shake????

    A slightly different way of looking at it: each of the n people shakes hands with each of the n- 1 hand shakes. So n people have n(n- 1) hand shakes. But each hand shake involves 2 people so the actual number of hand shakes is n(n- 1)/2. Solve n(n-1)/2= 66
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  6. #6
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    Re: Hand shake????

    Quote Originally Posted by mathmonster View Post
    basic drawn out method

    Triangle numbers . . .

    1 person shake hands 0 times (set n=0 as 1 person cannot create a handshake)
    2 people shake hands 1 time set n=1
    3 people shake hands 3 times set n=2
    4 people shake hands 6 times set n=3 etc

    n(n+1) = 66
    2

    n(n+1) = 132

    therefore you want two numbers with a difference of 1 that multiply together to give you 132.

    11x12=132 therefore n=11
    You are off a little. It is (n-1)(n)/2=66 or (n-1)(n)= 132=11*12. So n=12
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