If $\binom{n_1}{r}=\binom{n_2}{r}$,then is it true that $n_1=n_2?$ I know the identity that if $\binom{n}{r_1}=\binom{n}{r_2}$,then either $r_1=r_2$ or $r_1=n-r_2$.
But i want to ask if $\binom{n_1}{r}=\binom{n_2}{r}$,then is it true that $n_1=n_2?$.I am not sure if this is true or not.Please help.
 A: The identity $\binom{n+1}r=\binom nr+\binom n{r-1}$ should easily tell you for which values of $r\ge 1$ the function $n\mapsto \binom nr$ is strictly increasing.
A: Sorry it is not a complete answer but i hope it can help you .
p=>q  $\,\,\,\,\,\,\,\,$   ~q=>~p
p:=$\binom{n_{1}}r= \binom {n_{2}}r$
q:= $n_{1} = n_{2}$
we prove :
if ~q := $n_{1} \neq n_{2}$
then ~p:= $\binom{n_{1}}r\neq \binom {n_{2}}r$

$\binom{n+1}r-\binom nr=\binom n{r-1}$

$n_{1}:=n+1$
$n_{2}:=n$
if ~p is false them p is true and if P is true we have
$\binom{n_{1}}r= \binom {n_{2}}r$
and from: $\binom{n+1}r-\binom nr$= $\binom{n_{1}}r- \binom {n_{2}}r$$=\binom n{r-1}$
$\binom{n_{1}}r-\binom {n_{2}}r=\binom n{r-1}=0$
so $\binom n{r-1}=0$
that is always false
so we show that ~q=>~p
and find p=>q
A: For $n\ge r\ge1$, $\binom{n}{r}-\binom{n-1}{r}=\binom{n-1}{r-1}\gt0$.
However, $\binom{-n}{r}=(-1)^r\binom{r+n-1}{r}$ says that for even $r$, there may be two solutions to $\binom{n_1}{r}=\binom{n_2}{r}$.
For example,
$$
\binom{5}{2}=\binom{-4}{2}
$$
