TITLE: Why ${n \choose k} = {n \choose n-k}$?
QUESTION [13 upvotes]: 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:
$C(5,3)\times P(21,3)\times P(5,2)$

REPLY [2 votes]: 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.