 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability/discrete-mathematics/1/a/iii/solution

k

choices, with

n_{i}

i

regardless of earlier choices, has $\prod_{i = 1}^{k} n_{i}$ possible outcomes. Applied to the Cartesian product of finite sets, it gives

∣ A \times B ∣ = ∣ A ∣∣ B ∣

. Here independence means that the available choices do not restrict one another; no probability assumption is needed.

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