# Bijective proof

In combinatorics, bijective proof is a proof technique that finds a bijective function (that is, a one-to-one and onto function) f : AB between two finite sets A and B, or a size-preserving bijective function between two combinatorial classes, thus proving that they have the same number of elements, |A| = |B|. One place the technique is useful is where we wish to know the size of A, but can find no direct way of counting its elements. By establishing a bijection from A to some B solves the problem if B is more easily countable. Another useful feature of the technique is that the nature of the bijection itself often provides powerful insights into each or both of the sets.

## Basic examples

### Proving the symmetry of the binomial coefficients

The symmetry of the binomial coefficients states that

${\displaystyle {n \choose k}={n \choose n-k}.}$

This means that there are exactly as many combinations of k things in a set of size n as there are combinations of n − k things in a set of size n.

#### A bijective proof

The key idea of the proof may be understood from a simple example: selecting out of a group of n children which k to reward with ice cream cones has exactly the same effect as choosing instead the n − k children to be denied them.

More abstractly and generally,[1] the two quantities asserted to be equal count the subsets of size k and n − k, respectively, of any n-element set S. Let A be the set of all k-element subsets of S, the set A has size ${\displaystyle {\tbinom {n}{k}}.}$  Let B be the set of all n−k subsets of S, the set B has size ${\displaystyle {\tbinom {n}{n-k}}}$ . There is a simple bijection between the two two sets A and B: it associates every k-element subset (that is, a member of A) with its complement, which contains precisely the remaining n − k elements of S, and hence is a member of B. More formally, this can be written using functional notation as, f : AB defined by f(X) = Xc for X any k-element subset of S and the complement taken in S. To show that f is a bijection, first assume that f(X1) = f(X2), that is to say, X1c = X2c. Take the complements of each side (in S), using the fact that the complement of a complement of a set is the original set, to obtain X1 = X2. This shows that f is one-to-one. Now take any n−k-element subset of S in B, say Y. Its complement in S, Yc, is a k-element subset, and so, an element of A. Since f(Yc) = (Yc)c = Y, f is also onto and thus a bijection. The result now follows since the existence of a bijection between these finite sets shows that they have the same size, that is, ${\displaystyle {\tbinom {n}{k}}={\tbinom {n}{n-k}}}$ .

## Other examples

Problems that admit bijective proofs are not limited to binomial coefficient identities. As the complexity of the problem increases, a bijective proof can become very sophisticated. This technique is particularly useful in areas of discrete mathematics such as combinatorics, graph theory, and number theory.

The most classical examples of bijective proofs in combinatorics include: