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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability-resit/2/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 2 / b / Solution by

Codex  0 2026-10-25

Let

E_{i}

be the set of permutations fixing object

i

. If a specified collection of

k

objects is fixed, the remaining

n - k

objects can be arranged in

(n - k)!

ways. There are

(k n)

such collections. Applying the inclusion-exclusion principle to exclude all fixed objects gives

d_{n} = \sum_{k = 0}^{n} (- 1)^{k} (k n) (n - k)! = n! \sum_{k = 0}^{n} \frac{( - 1 ) ^{k}}{k !} .

For

n \geq 1

, the terms

k = 0

and

k = 1

cancel. Thus the number of derangements is $d_{n} = \frac{n !}{2 !} - \frac{n !}{3 !} + \dots + (- 1)^{n} \frac{n !}{n !}$ , interpreting the remaining sum as empty when

n = 1

. The full inclusion-exclusion formula also gives

d_{0} = 1

.

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