Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 17 / d / Solution

Under a uniform shuffle, every unordered five-card hand is equally likely, giving $\left(\genfrac{}{}{0ex}{}{52}{5}\right)$ possible hands. Choose the triple's rank in 13 ways and its suits in $\left(\genfrac{}{}{0ex}{}{4}{3}\right)$ ways; then choose a different pair rank in 12 ways and its suits in $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ ways. Each full house determines these choices uniquely, so there is no further ordering factor. ThereforeThe count uses binomial coefficients because the order of cards within a hand is irrelevant.