= Solution
A uniformly shuffled deck makes all unordered five-card hands equally likely, giving $\binom{52}{5}$ possible hands. Choose the rank appearing four times in $13$ ways; all four suits of that rank are then fixed. The remaining card can be any of the other $48$ cards. No hand is counted twice, because five cards cannot contain two different complete four-card ranks. This use of <combinatorics> gives the <probability>
$$
\boxed{\mathbb P(\text{four of a kind})=\frac{13\cdot48}{\binom{52}{5}}
=\frac{624}{\binom{52}{5}},\qquad N=624.}
$$
Here $\binom{52}{5}=2\,598\,960$ is a <binomial coefficient>; the <probability> also simplifies to $1/4165\approx0.000240096$.
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