Symmetry Rule for Binomial Coefficients/Proof 2

Theorem

 * $\displaystyle \forall n \in \Z, n > 0: \forall k \in \Z: \binom n k = \binom n {n - k}$

where $\displaystyle \binom n k$ is a binomial coefficient.

Proof
From the definition of Cardinality of Set of Subsets, $\displaystyle \binom n k$ is the number of subsets of cardinality $k$ of a set with cardinality $n$.

Let $S$ be a set with cardinality $n$.

Let $\mathcal P \left({S}\right)$ be the power set of $S$.

Let $\mathcal A_k \subseteq \mathcal P \left({S}\right)$ be the elements of $\mathcal P \left({S}\right)$ which have cardinality $k$.

For every $T \in \mathcal A_k$ there exists its relative complement $\complement_S \left({T}\right) \in \mathcal A_{n-k}$

Define the mapping $f_k: \mathcal A_k \to \mathcal A_{n-k}$ as:
 * $\forall T \in \mathcal A_k: f \left({T}\right) = \complement_S \left({T}\right)$

From Correspondence between Subset and Relative Complement it follows that $f_k$ is a bijection.

By definition of set equivalence, that means $\left|{A_k}\right| = \left|{A_{n-k}}\right|$.

That is, from the definition of Cardinality of Set of Subsets:
 * $\displaystyle \binom n k = \binom n {n - k}$