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If (n+4)!/(n+1)! = 990, what is the value of n?

Answer and explanation

Correct answer: 7

Answer: 7, so option A is correct. Expand the numerator to the denominator: (n+4)!=(n+4)(n+3)(n+2)(n+1)!. Cancelling (n+1)! gives (n+4)(n+3)(n+2)=990. The three factors are consecutive. Since 990=9×10×11, match n+2=9, n+3=10 and n+4=11. Therefore n=7. Check in the original expression: 11!/8!=11×10×9=990. Option B, n=8, gives 12×11×10=1320; option C gives 13×12×11=1716; and option D gives 14×13×12=2184. Thus none of them works. A useful cue is to arrange the remaining factors in increasing order; the smallest one is n+2, not n itself.

Related tags

Factorial EquationsConsecutive FactorsFactorial NotationPermutations And CombinationsMathematicsClass 11 Mcq

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