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There are \(4^5\) functions all together; we will subtract the functions which are not surjective. We could exclude any one of the four elements of the codomain, and doing so will leave us with \(3^5\) functions for each excluded element. This counts too many so we subtract the functions which exclude two of the four elements of the codomain, each pair giving \(2^5\) functions. But this excludes too many, so we add back in the functions which exclude three of the four elements of the codomain, each triple giving \(1^5\) function. There are \({4 \choose 1}\) groups of functions excluding a single element, \({4 \choose 2}\) groups of functions excluding a pair of elements, and \({4 \choose 3}\) groups of functions excluding a triple of elements. This means that the number of functions which are not surjective is:

\begin{equation*} {4 \choose 1}3^5 - {4 \choose 2}2^5 + {4 \choose 3}1^5\text{.} \end{equation*}

We can now say that the number of functions which are surjective is:

\begin{equation*} 4^5 - \left[{4 \choose 1}3^5 - {4 \choose 2}2^5 + {4 \choose 3}1^5\right]\text{.} \end{equation*}

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