 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/discrete-mathematics-continuous-mathematics-and-probability/2/b/solution

There are

n (n - 1) \dots (n - k + 1) = n! / (n - k)!

ordered selections of

k

k

subset appears in exactly

k!

(k n) = \frac{n !}{k ! ( n - k )!}, 0 \leq k \leq n .

For

k = 0

, both the empty selection and

0!

contribute one.

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