Show that $\binom{n}{r}=\binom{n}{n-r}$ by using the Binomial Formula Problem: Use the Binomial Formula to show that if $n$ and $r$ are integers with $0 \leq r \leq n$, then $\binom{n}{r}=\binom{n}{n-r}$.
My attempt: I am using the general binomial expansion formula to establish the following.
$(n+r)^{n}=n^{n}+nn^{n-1}r + ...$
But am not sure where to go from here. Should I do a proof by induction with base case $n=0$?  I have a feeling induction is not necessary and this problem is easier than I think...
Edit: This question is different than the proposed duplicate because the "duplicate" does not use the binomial theorem method, which I must use. It uses a different method.

Hint:
The polynomials $(x+y)^n$ and $(y+x)^n$ are the same. Expand and equal their coefficients.
Full answer:
The polynomials $(x+y)^n$ and $(y+x)^n$ are equal. If we apply the Binomial Formula to both of them, we obtain
$$\sum_{k=0}^n\binom nk x^ky^{n-k}=\sum_{j=0}^n\binom nj y^jx^{n-j}$$
Fix a degree $r$ for the variable $y$. The term that has this degree in the LHS is
$$\binom n {n-r}x^{n-r}y^r$$
and in RHS we find
$$\binom nr y^rx^{n-r}$$
This implies $$\binom nr=\binom n{n-r}$$
