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How does one find all non-negative integers $n,k$ such that ${n \choose k}=143$?

I factorized into $143=11 \cdot 13$, which means that $11 \cdot 13=\frac{n!}{k!(n-k)!}$, which implies that $n!=11 \cdot 13 \cdot k!(n-k)!$. This means that $13 \mid n!$, and if one thinks about the definition of factorial for a while, and due to the fact that $13$ is prime, we see that this implies $n \geq 13$. I'm stuck here though. Any ideas?

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    $\begingroup$ ${143\choose 1}=143={143\choose 142}$. $\endgroup$
    – vadim123
    Nov 27 '14 at 20:53
  • $\begingroup$ The possible biggest $n$ is clearly $143$. Now try everything below that (kidding). $\endgroup$ Nov 27 '14 at 20:54
  • $\begingroup$ @vadim123: I'm fully aware of that. What I'm looking for is a way to find all solutions and proving they are the only solutions. $\endgroup$
    – user195242
    Nov 27 '14 at 20:56
  • $\begingroup$ look in pascal triangle to find 143 , then you will have the combination $\endgroup$
    – Khosrotash
    Nov 27 '14 at 21:00
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    $\begingroup$ Except for $n=143$, we need the "small" $k$ to be $\ge 2$. Note that $\binom{20}{2}$ is already too big. Because of the $11$ we then need the small $k$ to be $\ge 3$. Then $\binom{13}{3}$ is already too big. $\endgroup$ Nov 27 '14 at 21:16
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It is easiest to assume by symmetry that $k \le n-k$. We have already seen that for $k=1, {143 \choose 1}=143$. If $k=2,$ we need $\frac 12n(n-1)=143$ and you can't get factors of both $13$ and $11$ for $n \lt 143$, so there are no solutions. For $n=3$, we need $\frac 16n(n-1)(n-2)=143.$ We can get the factors $11,13$ we want at $n=13$, but ${13 \choose 3}=286$ Similarly $k \ge 4$ fails, so we have them all.

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