2020/21 AHSMC Part 1 Problem 10

There are $\dbinom{10}{2}$ ways to choose a group of two from the $10$ competitors, then $\dbinom{8}{2}$ ways to choose a second group of two from the remaining eight, then $\dbinom{6}{2}$ ways to choose the third group of two from the remaining six. This leaves just four competitors left, who form the group of four. The order doesn't matter, so we must divide by $3!$ to eliminate duplications, and the requested number of choices is $$\frac{1}{3!} \binom{10}{2} \binom{8}{2} \binom{6}{2} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5}{6 \cdot 2 \cdot 2 \cdot 2} = 3150.$$ The answer is (B).