Combinatorial proof of $\binom{n}{k} = \binom{n}{n-k}$ How do I prove this combinatorially?

$$\displaystyle \binom{n}{k} = \binom{n}{n-k}$$

 A: If $X$ is a set, we denote its cardinality with $\# X$ . If $A$ is a set with $\# A = n$ and $0 \leq k \leq n$, then there exists a bijective mapping between the two following sets:
$$
U = \{ X \subseteq A \mid \ \#X = k \},
$$
$$
V = \{ Y \subseteq A \mid \ \#Y = n-k \};
$$
the map is $X \mapsto A \setminus X$.
A: Picking the $k$ elements you want out of $n$ possibilities amounts to the same thing as picking the $n-k$ elements you don't want out of $n$ possibilities.
A: $$
\binom 62 =15 = \binom 64.
$$
The number of ways to choose 2 out of 6 equals the number of ways to choose 4 out of 6, since every way of choosing 2 out of 6 corresponds to a way of choosing 4 out of 6, namely the 4 that are not among the chosen 2:
$$
\begin{array}{rcl}
AB & \leftrightarrow & CDEF \\
AC & \leftrightarrow & BDEF \\
AD & \leftrightarrow & BCEF \\
AE & \leftrightarrow & BCDF \\
AF & \leftrightarrow & BCDE \\
BC & \leftrightarrow & ADEF \\
BD & \leftrightarrow & ACEF \\
BE & \leftrightarrow & ACDF \\
BF & \leftrightarrow & ACDE \\
CD & \leftrightarrow & ABEF \\
CE & \leftrightarrow & ABDF \\
CF & \leftrightarrow & ABDE \\
DE & \leftrightarrow & ABCD \\
DF & \leftrightarrow & ABCE \\
EF & \leftrightarrow & ABCD
\end{array}
$$
(And simlilarly with other numbers.)
