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Math Help - (n,k) in binomial coefficients

  1. #1
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    (n,k) in binomial coefficients

    Calculation of integer ordered pair,s (n,k) for which \displaystyle \binom{n}{k} = 2002

    My approach:: Clearly (n,k)  = (2002,0)\;\;,(2002,2001) are the solution. Now we will find whether any ordered pair, exists OR not

    \displaystyle \binom{n}{k} = 2002  =  2 \times 7 \times 11 \times 13 = \frac{14 \times 13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}

    \displaystyle \binom{n}{k} = \binom{14}{9} = \binom{14}{5}

    So (n,k) = (14,9)\;\;,(14,5)

    But I did not understand How can I calculate any other pairs exists or not

    Help me please

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    Last edited by jacks; October 14th 2013 at 07:35 PM.
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  2. #2
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    Re: (n,k) in binomial coefficients

    I will give you two of them: \binom{2002}{1} = \binom{2002}{2001} = 2002.
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  3. #3
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    Re: (n,k) in binomial coefficients

    In general, this problem is unsolved. You can look at this link: (here). It implies that the only four solutions are (2002,1), (2002,2001), (14,9), (14,5).
    Thanks from jacks
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