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?

  • 1
    $\begingroup$ ${143\choose 1}=143={143\choose 142}$. $\endgroup$
    – 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.