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Math Help - Show equality by combinational argument

  1. #1
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    Show equality by combinational argument

    Hello,
    I came across a problem, and I was confused on how to go about doing it, so was seeing if I could get some advice. I guess I'm a little confused to whats going on with it, and not sure how to show by combinational arguments, and algebraic manipulation. Any help is greatly appreciated, thanks!
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  2. #2
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    Hello Mahonroy
    Quote Originally Posted by Mahonroy View Post
    Hello,
    I came across a problem, and I was confused on how to go about doing it, so was seeing if I could get some advice. I guess I'm a little confused to whats going on with it, and not sure how to show by combinational arguments, and algebraic manipulation. Any help is greatly appreciated, thanks!
    (a) \binom{2n}{2} is the number of ways of choosing 2 items from a group containing 2n items.

    Now we may imagine the 2n items to be divided arbitrarily into two groups of n items each, and we may choose:
    • 2 items from the first group (which can be done in \binom{n}{2} ways); or

    • 2 items from the second group (again in \binom{n}{2} ways); or

    • 1 item from each group (which can be done in n \times n = n^2 ways).


    Thus the total number of ways of carrying out one of these three options is \binom{n}{2}+\binom{n}{2}+n^2=2\binom{n}{2}+n^2

    Thus \binom{2n}{2}=2\binom{n}{2}+n^2


    (b) \binom{2n}{2} = \frac{2n!}{2!(2n-2)!}= \frac{2n(2n-1)}{2}= n(2n-1)

    2\binom{n}{2}+n^2=2\cdot\frac{n!}{2!(n-2)!}+n^2=\frac{2n(n-1)}{2}+n^2

     = n(n-1)+n^2 = n(2n-1) = \binom{2n}{2}

    Grandad
    Last edited by Grandad; September 30th 2009 at 02:23 PM. Reason: More complete solution
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  3. #3
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    Ok this is making more sense to me now thanks!
    While sitting down earlier I figured of using the (n / r) = n!/(r!*(n-r)!), so I just placed those values in there like that and simplified both sides to get n(2n-1)=n(2n-1), and since I used the combinational theorems for it, I thought that it would maybe go under the "by a combinational argument", and at that point I was confused about the "by an algebraic manipulation", because it seemed that that way is kind of a little of both, and I didn't really see any other way of proving it really. So that is really considered the algebraic way, and the first part that you were explaining is the combinational way? Thanks again
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