Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 14 / a

By the multiplication principle, the successive places have six, five, four, three, two and one available letters. Thus the number of permutations is

Fixing the last letter leaves five distinct letters to permute freely. Therefore the number is

The last letter has two permitted choices, and each leaves $5!$ permutations. The two classes are disjoint, so the count is

Fixing both specified positions leaves four letters to permute. Hence

Let $E_5$ denote the set of arrangements with $E$ fifth, and $F_6$ those with $F$ sixth. Each contains $5!$ arrangements, while their intersection contains $4!$. The inclusion-exclusion principle gives

Requiring exactly one condition means taking the symmetric difference of the two sets. Subtract their shared cases twice from the two individual counts:Equivalently, remove the twenty-four arrangements satisfying both conditions from the inclusive count $216$.