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If you happen to calculate this number precisely, you will get 120 surjections. That happens to also be the value of \(5!\text{.}\) This might seem like an amazing coincidence until you realize that every surjective function \(f:X \to Y\) with \(\card{X} = \card{Y}\) finite must necessarily be a bijection. The number of bijections is always \(\card{X}!\) in this case. What we have here is a combinatorial proof of the following identity:

\begin{equation*} n^n - \left[{n\choose 1}(n-1)^n - {n \choose 2}(n-2)^n + \cdots + {n \choose n-1}1^n \right] = n!\text{.} \end{equation*}

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