1996 AHSME Problems/Problem 3

Problem

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

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

Solution

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

The denominator is$ (Error compiling LaTeX. Unknown error_msg)3! = 6$.

Using the property that$ (Error compiling LaTeX. Unknown error_msg)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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1996 AHSME Problems/Problem 3

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