= Solution
Under a uniform shuffle, every unordered five-card hand is equally likely, giving $\binom{52}{5}$ possible hands. Choose the triple's rank in 13 ways and its suits in $\binom43$ ways; then choose a different pair rank in 12 ways and its suits in $\binom42$ ways. Each full house determines these choices uniquely, so there is no further ordering factor. Therefore
$$
\boxed{N=13\binom43\,12\binom42=3744,\qquad P=\frac{3744}{\binom{52}{5}}.}
$$
The count uses <binomial coefficients> because the order of cards within a hand is irrelevant.
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