Why ${n \choose k} = {n \choose n-k}$?
Solution 1:
$$ \begin{align} \dbinom{6}{2} & \longleftrightarrow \dbinom{6}{6-2} \\[8pt] AB & \longleftrightarrow CDEF \\ AC & \longleftrightarrow BDEF \\ AD & \longleftrightarrow BCEF \\ AE & \longleftrightarrow BCDF \\ AF & \longleftrightarrow BCDE \\ BC & \longleftrightarrow ADEF \\ BD & \longleftrightarrow ACEF \\ BE & \longleftrightarrow ACDF \\ BF & \longleftrightarrow ACDE \\ CD & \longleftrightarrow ABEF \\ CE & \longleftrightarrow ABDF \\ CF & \longleftrightarrow ABDE \\ DE & \longleftrightarrow ABCF \\ DF & \longleftrightarrow ABCE \\ EF & \longleftrightarrow ABCD \end{align} $$ There are exactly as many ways to choose $2$ out of $6$ as to choose $6-2$ out of $6$ because each way of choosing $2$ out of $6$ has a corresponding way of choosing $6-2$ out of $6$ and vice-versa.
Solution 2:
Consider a collection of $n$ objects. Choosing $k$ of them to place into a set is equivalent to choosing $n-k$ to leave out.
Edit: Consider a high school dodgeball game with a red team and a blue team. There are $n$ total students, and the blue team has $k$ students. Since every student plays, there are $n-k$ students on the red team.
Because the PE teacher is biased, he lets the blue team pick all of their players first. They have $\binom{n}{k}$ ways to do this. After that, the red team has no choices. They must pick all of the remaining $n-k$ students.
There would be an equal number of ways to do this if the teacher was biased towards red, so $\binom{n}{k} = \binom{n}{n-k}$.
Solution 3:
When we have $n$ objects to choose from, and we choose to include $k$ of them, there are ${n\choose k}$ ways of choosing these objects. However, at the same time we are choosing not to include $n-k$ objects, and there are ${n \choose n-k}$ ways of excluding these objects. Thus we have that ${n \choose k} = {n \choose n-k}$.
Solution 4:
Symmetry of summation.
$$ (x+y)^n - (y+x)^n = 0, $$
so
$$ \sum_{k=0}^n {n \choose k} x^k y^{n-k} - \sum_{k=0}^n {n \choose k} y^k x^{n-k} = 0, $$
whence
$$ \sum_{k=0}^n \left[ {n \choose k} - {n \choose n-k} \right] x^k y^{n-k} = 0, $$
as valid for aribtrary $x$ and $y$ we obtain
$$ {n \choose k} = {n \choose n-k} $$
The physical meaning - indeed...
Given are $(k)$ white balls and $(n-k)$ black balls, we can form $\displaystyle n \choose \displaystyle k$ permutations.
Given are $(n-k)$ white balls and $(k)$ black balls, we can form $\displaystyle n \choose \displaystyle n-k$ permutations.
Both cases are symmetrical (black ball $\leftrightarrow$ white ball), so
$$ {n \choose k} = {n \choose n-k } $$
Solution 5:
You can use the definition of $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ :
$\binom{n}{n-k}=\frac{n!}{(n-k)!(n-(n-k))!}=\frac{n!}{(n-k)!(n-n+k)!}=\frac{n!}{(n-k)!(k)!}=\frac{n!}{k!(n-k)!}=\binom{n}{k}$
In my experience this is the easiest proof that I found of this without having to go through a lot of combinatorial arguments. It just "drops out" from algebraic manipulation due to the inherit symmetry.