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Thread: (n,k) in binomial coefficients

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

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

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

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

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

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

    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; Oct 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: $\displaystyle \binom{2002}{1} = \binom{2002}{2001} = 2002$.
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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 $\displaystyle (2002,1), (2002,2001), (14,9), (14,5)$.
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