Finding all ordered pair of natural number $(n,r)$

for which $\displaystyle \binom{n}{r} = 240$

Try: $$\binom{n}{r}=\binom{n}{n-r}=240$$

For $n=240$ and $r=1.$ we have $\displaystyle \binom{240}{1}=240$

For $n=240$ and $r=239.$ we have $\displaystyle \binom{240}{239}=240$

If $n\in $ Even natural number. then $\displaystyle \binom{n}{r}$ is maximum for $r=n/2$

If $n\in $ Odd natural number. then $\displaystyle \binom{n}{r}$ is maximum for $r=(n+1)/2$ and $r=(n+3)/2$

not know how to solve ahead from that point

Could some explain me how i find other ordered pair of natural number

  • 1
    $\begingroup$ Suppose $n\choose2$ is 240. That gives you an equation for $n$, which you can try to solve. Similarly for $n\choose3$, $n\choose4$, and so on. That's a good start. Why don't you try it, and see how far you get, and report back. $\endgroup$ – Gerry Myerson Oct 16 '18 at 11:03
  • $\begingroup$ @Gerry Myerson i have a problem that how can i check for all such values like $\binom{n}{2}=240,\binom{n}{3}=240,\binom{n}{4}=240......$. would you explain me in detail .thanks $\endgroup$ – DXT Oct 16 '18 at 14:16
  • 1
    $\begingroup$ The answer that has been posted shows you what you could have done if you had tried a little harder instead of asking others to do the work for you. $\endgroup$ – Gerry Myerson Oct 16 '18 at 21:35

Let us consider the cases where $r\le \lfloor\frac n2\rfloor$.

  • For $n\le 9$, we have $$\binom nr\le\binom{9}{4}=126\lt 240$$

  • For $n=10$, we get $$\binom{10}{4}=210\lt 240\lt 252=\binom{10}{5}$$

  • For $n\ge 11$ and $4\le r\le\lfloor \frac n2\rfloor$, we have $$\binom nr\ge\binom{11}{4}=330\gt 240$$

  • For $r=1$, we have $n=240$.

  • For $r=2$, the equation is equivalent to $$\frac{n(n-1)}{2}=240,$$i.e.$$n(n-1)=480$$But there are no such $n$ since $$22\times 21=462\lt 480\lt 506=23\times 22$$

  • For $r=3$, the equation is equivalent to $$\frac{n(n-1)(n-2)}{6}=240,$$i.e.$$n(n-1)(n-2)=1440$$But there are no such $n$ since$$12\times 11\times 10=1320\lt 1440\lt 1716=13\times 12\times 11$$

Conclusion :

The only pairs $(n,r)$ such that $\binom nr=240$ are $$(n,r)=(240,1),(240,239)$$


Solution with brute force:

The enclosed figure presents a part of the right half of first rows of Pascal's triangle. Starting with the 23rd row we are sure that only $\binom{n}{1}$ (and $\binom{n}{n-1}$ by symmetry) can give $240.$ This occurs for $n=240.$ There exist no other solution, only those you have found.PT


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