1996 AHSME Problems/Problem 3

Problem

$\frac{(3!)!}{3!}=$

$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120$

The numerator is $(3!)! = 6!$ .

The denominator is $3! = 6$ .

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

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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