2024/25 AHSMC Part 1 Problem 5

Temporarily label the boxes so they are distinct. There are $$\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{10!}{2^5}$$ ways to distribute the cookies. Removing the labels, every distribution is counted $5!$ times, so the number of ways is $$\frac{10!}{2^5\,5!} = 945.$$ Alternatively: after placing the first cookie in a box there are $9$ choices for its partner, then $7$ for the next pair, then $5$, then $3$, giving $9 \times 7 \times 5 \times 3 = 945$. The answer is (A).