# Difference between revisions of "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 $3! = 6$.

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