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$(1)$ Evaluation of all positive integer ordered pair $(n,r)$ for which $\displaystyle \binom{n}{r} = 120$

$(2)$ Evaluation of all positive integer ordered pair $(n,r)$ for which $\displaystyle \binom{n}{r} = 2016$

$\bf{My\; Try::}$ Here $r=1$ and $n=120\;,$ Then $\displaystyle \binom{120}{1} = 120$

So $\displaystyle (n,r) = (120,1)\;\;,(120,119)$ (Using $\displaystyle \binom{n}{r} = \binom{n}{n-r}$)

Here $\displaystyle \binom{9}{3} = 84$ and $\displaystyle \binom{9}{4} = 126.$

So using Triangular numbers, Here $\displaystyle \binom{n}{r} = 120$ is valid, when $n>9$

Now when $r=2$ and $n=16\;,$ Then $\displaystyle \binom{16}{2} = 120$

So $\displaystyle (n,r) = (16,2)\;\;,(16,14)$

Now we will find when $\displaystyle \binom{n}{r}$ is an Increasing function.

So $\displaystyle \frac{\binom{n}{r+1}}{\binom{n}{r}}\geq 1\Rightarrow \frac{n-r}{r+1}\geq 1\Rightarrow r\leq \frac{n-1}{2}$

So If $r\geq 3\;,$ and $n>9\;,$ Then $\displaystyle 120=\binom{n}{r}>\binom{n}{3} = \frac{n(n-1)(n-2)}{6}>\frac{(n-2)^3}{6}$

So $\displaystyle (n-2)^3<720<(9)^3\Rightarrow n<11$

So we have $n=10$ and $r=3\;,$ So we get $\displaystyle \binom{10}{3} = \binom{10}{7} = 120$

So we get $\displaystyle (n,r) = (10,3) = (10,7)$

So we get Total positive integer ordered pairs $$\displaystyle (m,n) = \left\{(120,1)\;\;,(120,119)\;\;,(16,2)\;\;,(16,8)\;\;,(10,3)\;\;,(10,7)\right\}$$

But i did not understand How can I solve $(2)$ one.

Although we know that $\displaystyle \binom{2016}{1} = \binom{2016}{2015} = 2016$

Help me, Thanks

Marius S.L.
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juantheron
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  • If $n > 2016$, there are no solutions to $\binom{n}{k}=2016$. This makes your search finite. – marty cohen Jan 31 '16 at 05:52
  • Here is a hint for 2016. As you noted, since $\binom{n}{r} = \binom{n}{n-r}$, it's enough for look for solutions with $0 \leq r \leq n/2$. Within this range, $\binom{n}{r}$ increases when $r$ increases. We have $\binom{24}{3} > 2016$, so the only possibilities with $r \geq 3$ will have $n \leq 23$. Also, for a fixed value of $r$, $\binom{n}{r}$ increases with $n$ ($n\geq r$), so there will be at most one value of $n$ that works for that $r$. Finally, since $7 | 2016$, you can narrow the possibilities by examining how many times $7$ appears in the numerator and denominator of $\binom{n}{r}$. – David Jan 31 '16 at 07:17
  • Thanks Marty, Thanks David , but i did not understand how can i found $\displaystyle \binom{64}{2} = \binom{64}{62} = 2016$ – juantheron Jan 31 '16 at 07:57
  • Is this a question from an on-going contest? – JRN Mar 20 '16 at 13:29
  • I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – GNUSupporter 8964民主女神 地下教會 Feb 27 '19 at 09:59

2 Answers2

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I wrote a simple program to search all $\binom{n}{k}$ for $n < 2016$.

The only solutions were the ones you found.

For a more sophisticated search, I could use the factorization $2016=2^53^27$, to limit the possible values of $n$ and $k$, but not now.

marty cohen
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2

(2)

The fact that $\displaystyle\binom nr\leqslant \binom{n}{\lfloor\frac{n}{2}\rfloor}$ should be helpful.

If $n\leqslant 13$, there are no solutions because $\binom nr\leqslant\binom{13}{6}=1716\lt 2016$.

If $n=14$, there are no solutions because $\binom{14}{5}=2002\lt 2016\lt 3003=\binom{14}{6}$.

In the following, $n\geqslant 15$.

  • For $r=1$, we have $n=2016$.

  • For $r=2$, solving $\frac{n(n-1)}{2}=2016$ gives $n=64$.

  • For $r=3$, we have $n(n-1)(n-2)=2^63^37^1$. Since LHS has only one integer divisible by $3$, LHS has an integer divisible by $27$. So, we have to have $n\geqslant 27$, but then $\binom{n}{3}\geqslant\binom{27}{3}=\frac{27\cdot 26\cdot 25}{6}=9\cdot 13\cdot 25>8\cdot 12\cdot 25=2400\gt 2016$.

  • For $r=4$, we have $n(n-1)(n-2)(n-3)=2^83^37^1$. Since LHS has only one integer divisible by $4$, LHS has an integer divisible by $32$. So, we have to have $n\geqslant 32$, but then $\binom n4\geqslant\binom{32}{4}=\frac{32\cdot 31\cdot 30\cdot 29}{24}\gt \frac{32\cdot 31\cdot 30\cdot 29}{32}\gt 30\cdot 30\cdot 10=9000\gt 2016$.

  • For $5\leqslant r\leqslant\lfloor \frac n2\rfloor$, we have $\binom nr\geqslant\binom{15}{5}=3003\gt 2016$.

Therefore, the answer is $$\color{red}{(n,r)=(2016,1),(2016,2015),(64,2),(64,62)}$$

mathlove
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