View source for 1996 AHSME Problems/Problem 3

==Problem==

<math> \frac{(3!)!}{3!}= </math>

<math> \text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120 </math>

==Solution==

The numerator is <math>(3!)! = 6!</math>.

The denominator is <math>3! = 6</math>.

Using the property that <math>6! = 6 \cdot 5!</math> in the numerator, the sixes cancel, leaving <math>5! = 120</math>, which is answer <math>\boxed{E}</math>.

==See also==
{{AHSME box|year=1996|num-b=2|num-a=4}}
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