Let ${{n} \choose {r}} = 8$.
Is there any other choice of $n$ and $r$ except $8$ and $1$, $8$ and $7$ ?

In general how to check that existence is guaranteed or not?

  • 7
    $\begingroup$ there's also $8$ and $7$ (these things are symmetric) $\endgroup$ – Gregory Grant May 7 '15 at 11:16
  • $\begingroup$ Ya you are right. Is there any other such pair. $\endgroup$ – Sry May 7 '15 at 11:18
  • $\begingroup$ I doubt it, 8 is a power of 2 and factorials are usually going involve other primes. That's not a proof though. $\endgroup$ – Gregory Grant May 7 '15 at 11:19
  • $\begingroup$ @Gregory Grant, How power of 2 is helping here? It may be but there are factorials in denominator as well. $\endgroup$ – Sry May 7 '15 at 11:32
  • $\begingroup$ What do you mean by "check that existence is guaranteed"? $\endgroup$ – Cameron Buie May 7 '15 at 11:42

$\binom{n}{r}\geq n$ as soon as $r\neq 0,n$. So the only way to have $\binom{n}{r}=8$ is for $n\leq 8$. A quick check (on a Pascal triangle) shows that the only solution is $\binom{8}{1}=\binom{8}{7}$


You must have $n\le 8$. If $n>8$, the smallest binomial coefficient $\binom nr$ other than $1$ (when $r=0$ or $n$) is $\binom n1=n$. So there are only a finite number of cases to check.

  • $\begingroup$ @ Tad, in case 8 is replaced by a large number, is there any other way to check the existence except checking all the finite possible cases.Though I should have asked it in the question itself. $\endgroup$ – Sry May 7 '15 at 11:29
  • $\begingroup$ The next option is ${m\choose2}={m(m-1)\over2}=n$, for which $8n+1=(2m-1)^2$. After that you only have to check for $m<\sqrt{2n}$ $\endgroup$ – Empy2 May 7 '15 at 11:49
  • $\begingroup$ Not sure what you mean by "all the finite possible cases." I would simply try each row in Pascal's triangle in order, starting from the left. You stop looking in the $m$-th row as soon as $\binom mr>n$, and you remember that $r$ for the next row, since $\binom {m+1}r$ will be even bigger. An asymptotic analysis along the lines @Michael suggests is possible, but I think is beyond the scope of what you're looking for. $\endgroup$ – Tad May 7 '15 at 12:02

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