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What is the value of (14!/(12!·2!) + 13!/(11!·2!))?

Answer and explanation

Correct answer: 169

The governing idea is cancellation in factorial notation. Since 14! = 14·13·12! and 2! = 2·1 = 2, the first term is 14·13/2 = 182/2 = 91. Similarly, 13! = 13·12·11!, so the second term is 13·12/2 = 156/2 = 78. Adding the two evaluated terms gives 91 + 78 = 169. Therefore option B is correct. A common error is to cancel only one factorial factor or to forget that 2! equals 2 rather than 1. Options A, C, and D result from incorrect cancellation or arithmetic, whereas 169 follows directly from expanding only the factors needed to cancel the denominators.

Related tags

Factorial NotationPermutations And CombinationsCancellationMathematicsClass 11 Mcq

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