 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/discrete-mathematics-continuous-mathematics-and-probability/1/c/iii/solution

{\emptyset}

has cardinality one, so its power set has cardinality two. Taking the Cartesian product with the singleton

{\emptyset}

leaves that cardinality equal to two. The outer power set therefore has $2^{2} = 4$ elements.

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