5
$\begingroup$

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?

$\endgroup$
6
  • 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
2
$\begingroup$

$\binom{n}{r}\geq n$ 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 $\binom{8}{1}=\binom{8}{7}$

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ @ 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

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.