What is ( \frac{(3n)!}{(3n-2)!} ) equal to?

Correct answer: (3n(3n-1))

The direct answer is option A: \(3n(3n-1)\). Factorial means multiplying all positive integers down to 1. Therefore \((3n)!=(3n)(3n-1)(3n-2)\cdots1\). Also, \((3n-2)!=(3n-2)(3n-3)\cdots1\). When dividing, every factor from \((3n-2)\) down to 1 cancels, leaving \(3n(3n-1)\). Option A is exactly this result. Option B incorrectly uses \(3n-2\) instead of \(3n-1\) as the second remaining factor. Option C leaves factors that belong to the denominator and misses the factor \(3n\). Option D treats the result as \((3n)^2\), but the two remaining consecutive factors are different, not both \(3n\). The rule works for values where the factorials are defined, ordinarily non-negative integer arguments. Memory cue: expand only the larger factorial until the smaller factorial appears, then cancel.