1996 AHSME Problems/Problem 3

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Problem

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

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

Solution

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

The denominator is$ (Error compiling LaTeX. ! Missing $ inserted.)3! = 6$.

Using the property that$ (Error compiling LaTeX. ! Missing $ inserted.)6! = 6 \cdot 5!$in the numerator, the sixes cancel, leaving$5! = 120$, which is answer$\boxed{E}$.

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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