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

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 $216$.

The second bin is forced to contain $A,B$, leaving the other four letters in the first bin. Since order within a bin is irrelevant, this gives exactlyset partition of the specified bin sizes.

The first four letters may be permuted in $4!$ ways and the last two in $2!$ ways without changing this partition. The multiplication principle therefore gives

Choose the unordered pair of letters in the second bin; the first bin is then determined. The binomial coefficient givesThe first and second bins are distinguished, so their roles are not interchangeable.

The product isEvery permutation determines exactly one bin partition, and each such partition has exactly forty-eight internal orderings. Thus the fifteen classes partition the original sample space of $720$ arrangements, without overlap or omission.

For first-bin size $k$, choose its contents in $\left(\genfrac{}{}{0ex}{}{6}{k}\right)$ ways. Each choice arises from $k!(6-k)!$ permutations, obtained by ordering the two bins internally. The seven ordered size choices giveHere $0!=1$. Each row expresses the same binomial coefficient identity $\left(\genfrac{}{}{0ex}{}{6}{k}\right)k!(6-k)!=6!$. Even for $\{3,3\}$ the bins retain their first/second labels, so the number is twenty rather than ten; the braces denote bin sizes in the question's convention, not interchangeable bins.