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Math Help - word problem.... very tricky

  1. #1
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    word problem.... very tricky

    can anyone explain how he got 28????
    Attached Thumbnails Attached Thumbnails word problem.... very tricky-screen-shot-2012-12-19-1.21.42-pm.png  
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  2. #2
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    Re: word problem.... very tricky

    count 'em ...



    or think ...

    2 computers ... 1 wire
    3 computers ... 3 wires
    4 computers ... 6 wires
    5 computers ... 10 wires

    see the pattern?
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  3. #3
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    Re: word problem.... very tricky

    Quote Originally Posted by asilvester635 View Post
    can anyone explain how he got 28????

    This is known in graph theory as a complete graph of order eight: \mathbb{K}_8.

    The number of edges in \mathbb{K}_n is \binom{n}{2}=\frac{n(n-1)}{2}.
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  4. #4
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    Re: word problem.... very tricky

    Yes, the sum of numbers up to n is (n*(n-1))/2; therefore, for the n=8, the answer is 56/2 = 28. We need one fewer wire, since we are not connecting a computer to itself, as explained in the answer already provided. We need that for all 8 computers, considered as a pair each time. Therefore 8*7 / 2 is both logical and correct.
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    Re: word problem.... very tricky

    Quote Originally Posted by cathectio View Post
    Yes, the sum of numbers up to n is (n*(n-1))/2; therefore, for the n=8, the answer is 56/2 = 28. We need one fewer wire, since we are not connecting a computer to itself, as explained in the answer already provided. We need that for all 8 computers, considered as a pair each time. Therefore 8*7 / 2 is both logical and correct.
    NO! That is absolutely not correct.
    The formula I gave you in no way implies that a vertex is connected to itself.
    This is such a well-known question, why do you question my reply?
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  6. #6
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    Re: word problem.... very tricky

    I reiterate what was posted: we need one fewer wire. I added that we are NOT connecting a computer to itself, as you seem to imply the opposite.
    If one draws out the connections, they come to 28. BUT THE LOGICAL ANSWER GIVEN, THAT WE NEED ONE FEWER WIRE FOR ALL 8 CONNECTIONS RESULTS IN THE FORMULA I GAVE, N(N-1)/2.
    What part of that do you have a problem with, Plato?
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