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If \( \frac{(3n)!}{(3n-2)!}=132 \), what is the value of (n)?

Answer and explanation

Correct answer: 4

Cancelling \((3n-2)!\) from \((3n)!\) gives \(3n(3n-1)=132\). Since \(12\times 11=132\), we get \(3n=12\). Therefore, \(n=4\). The alternative value \(3n=-11\) is invalid because factorials are not defined for negative integers. Exam tip: In factorial ratios, expand the larger factorial only up to the factors needed for cancellation.

Related tags

Factorial NotationFactorial SimplificationPermutations And CombinationsAlgebraic EquationsClass 11 MathematicsMathematicsClass 11 Mcq

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