 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/discrete-mathematics-continuous-mathematics-and-probability/1/d/solution

Let

x_{i}

be the number in box

i

and put

y_{i} = x_{i} - 1

y_{1} + \dots + y_{4} = 6

and

0 \leq y_{i} \leq 4

. Without the upper bounds, stars and bars gives

(3 9) = 84

. If a specified

y_{i} \geq 5

, subtract five from it; the remaining nonnegative integer variables sum to one, giving

(3 4) = 4

possibilities. Two violations cannot occur because they would require a sum at least ten. The inclusion-exclusion principle therefore gives

(3 9) - 4 (3 4) = 84 - 16 = 68.

The boxes are distinguished by their positions in the composition; individual objects are not distinguished.

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