# oreder pair of natural number in binomial coefficient

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

• 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. – Gerry Myerson Oct 16 '18 at 11:03
• @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 – DXT Oct 16 '18 at 14:16
• 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. – 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. 