 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/a/ii/solution

Put one object into each box first. The remaining

n - k

objects can be distributed without restrictions, so stars and bars gives

(k - 1 ( n - k ) + k - 1) = (k - 1 n - 1) .

Equivalently, the substitution

u_{i} = x_{i} - 1

gives a bijection between positive integer solutions of

\sum_{i} x_{i} = n

and nonnegative integer solutions of

\sum_{i} u_{i} = n - k

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