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:
$C(5,3)\times P(21,3)\times P(5,2)$
Best Answer
$$ \begin{align} \dbinom{6}{2} & \longleftrightarrow \dbinom{6}{6-2} \\[8pt] AB & \longleftrightarrow CDEF \\ AC & \longleftrightarrow BDEF \\ AD & \longleftrightarrow BCEF \\ AE & \longleftrightarrow BCDF \\ AF & \longleftrightarrow BCDE \\ BC & \longleftrightarrow ADEF \\ BD & \longleftrightarrow ACEF \\ BE & \longleftrightarrow ACDF \\ BF & \longleftrightarrow ACDE \\ CD & \longleftrightarrow ABEF \\ CE & \longleftrightarrow ABDF \\ CF & \longleftrightarrow ABDE \\ DE & \longleftrightarrow ABCF \\ DF & \longleftrightarrow ABCE \\ EF & \longleftrightarrow ABCD \end{align} $$ 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.