 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-2/4f/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

k

digits must be distinct and may be chosen in

(10)_{k} = 10 \cdot 9 \dots (10 - k + 1)

ways; the remaining digits are then forced in reverse order. Thus, for

k \leq 10

P (E) = \frac{( 10 ) _{k}}{1 0 ^{2 k}},

and the probability is zero for

k > 10

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