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 1

\begin{align*} \frac{6!}{6} &= \frac{\cancel{6} \cdot 5!}{\cancel{6}} \\ &= 5! = \boxed{\textbf{(E) }120} \end{align*}

Solution 2

\begin{align*} \frac{(3!)!}{3!} &= \frac{6!}{3!} \\ &= \frac{720}{6} \\ &= \boxed{\textbf{(E) }120} \end{align*}

See Also

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