In this discussion, a matching $M$ in a relation $R$ is simply a subset of $R$ where all pairs are distinct – thus for $(a,b) \in M$ and $(c,d) \in M$, $a \neq b$ and $b \neq d$. A perfect matching in $R$ has $|A|$ pairs. To avoid appearing repetitive, we will assume that any mention of a matching refers to $M \subseteq R$, where $R \subseteq A \times B$ is a relation from $A$ to $B$. When a matching is so special as to be perfect, we will be explicit about it (and thus a matching is not to be considered perfect by default). The Marriage Theorem states $R$ admits a perfect matching if, and only if,

$$\forall X \subseteq A, |R(X)| \geq X.$$

We will presently resist the temptation to describe any of the many proofs of this theorem, and proceed assuming the truth of the statement. Since the theorem gives us a characterization for relations that admit perfect matchings, it is natural to consider those that don’t. So, what is it about a perfect matching that makes it neat?

Let’s say a matching saturates an element $a \in A$ if $(a,x) \in M$ for some $x \in B$. A perfect matching clearly saturates all elements of $A$. As a first step towards looking at non-perfect matchings, let us try those that saturate not all, but simply “at least a certain number” of the elements in $A$. In particular, let’s say that a matching $M$ has deficit $d$ if it saturates all but $d$ elements in $A$.

$$C(n,k) = C(n-1,k-1) + C(n-1,k) (n-1),$$

where $C(n,k)$ denotes the number of permutations of, say, $[n]$ with $k$ cycles. The equality is nearly self-explanatory — you simply knock off one of the elements (let’s do away with $n$) and stare at permutations of $[n-1]$.

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