 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability-resit/2/c/solution

The initial values of the two sequences agree:

c_{0} = d_{0} = 1

and

c_{1} = - 1 + c_{0} = 0 = d_{1}

. For

n \geq 2

n

and

n - 1

c_{n} - (n - 1) (c_{n - 1} + c_{n - 2}) = (- 1)^{n} + c_{n - 1} - (n - 1) c_{n - 2} = (- 1)^{n} + (- 1)^{n - 1} = 0.

Therefore

c_{n}

satisfies exactly the same derangement recurrence as

d_{n}

. By strong induction, equality of the preceding two terms implies equality of the next term. Consequently

c_{n} = d_{n} for every n \geq 0.

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