<s>[INST] Evaluation of all positive integer ordered pair $(n,r)$ for which $\binom{n}{r} = 2016$ 
$(1)$ 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
 [/INST] 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.
</s>