Why ${n \choose k} = {n \choose n-k}$? They say that $${n \choose k}={n \choose n-k}.$$ 
Can someone explain its meaning?

Among many problems that use this proof, here is an example:

The english alphabet has $26$ letters of which $5$ are vowels (and $21$ are
  consonants).
How many $5$-letter words can we form by using $3$ different consonants
  and $2$ different vowels?

I understand where the answer says we have:
$$P(21,3) = 21\times 20\times 19 = 7980\ ,$$
and 
$$P(5,2) = 5\times4 = 20\ .$$
We get the permutations for each category. Now we must place them into $5$ places, but it says this is done by computing:
$$C(5,3)$$
and it explains further: 

For each case, the rest of the letters will be vowels.

(Aren't we supposed to check that case?)
It ends by multiplying all three together:
$C(5,3)\times P(21,3)\times P(5,2)$
 A: 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 ${n \choose k} = {n \choose n-k}$.
A: 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
$$
{n \choose k} =  {n \choose n-k }
$$
A: 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.
A: $$
\begin{align}
\dbinom{6}{2} & \longleftrightarrow  \dbinom{6}{6-2} \\[8pt]
AB & \longleftrightarrow CDEF \\
AC & \longleftrightarrow BDEF \\
AD & \longleftrightarrow BCEF \\
AE & \longleftrightarrow BCDF \\
AF & \longleftrightarrow BCDE \\
BC & \longleftrightarrow ADEF \\
BD & \longleftrightarrow ACEF \\
BE & \longleftrightarrow ACDF \\
BF & \longleftrightarrow ACDE \\
CD & \longleftrightarrow ABEF \\
CE & \longleftrightarrow ABDF \\
CF & \longleftrightarrow ABDE \\
DE & \longleftrightarrow ABCF \\
DF & \longleftrightarrow ABCE \\
EF & \longleftrightarrow ABCD 
\end{align}
$$
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.
A: 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 $\binom{n}{k} = \binom{n}{n-k}$.
A: If you have $n=a+b$ boxes of which $a$ contain a ball and $b$ are empty, you can choose $a$ to contain a ball in $\binom na$ ways or $b$ to be empty in $\binom nb$ ways. The two are clearly equivalent, because they are counting the same thing.
A: $\binom{n}{k}$ represents the cardinality of set of all subsets of cardinality $k$ from a set of cardinality $n$. The fact that $\binom{n}{k}=\binom{n}{n-k}$ represents that there are the same number of elements in the new set you obtain when you take the complement of each of the $k$ element subsets.
