Difference between revisions of "1996 AHSME Problems/Problem 3"

Latest revision as of 17:37, 26 June 2025

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

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

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

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

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
All AHSME Problems and Solutions

@@ Line 1: / Line 1: @@
-==See also==
+== Problem ==
-{{AHSME box|year=1995|num-b=7|num-a=9}}
+<cmath> \frac{(3!)!}{3!}= </cmath>
+<math> \text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120 </math>
+== Solution 1 ==
+<cmath>
+\begin{align*}
+\frac{6!}{6} &= \frac{\cancel{6} \cdot 5!}{\cancel{6}} \\
+&= 5! = \boxed{\textbf{(E) }120}
+\end{align*}
+</cmath>
+== Solution 2 ==
+<cmath>
+\begin{align*}
+\frac{(3!)!}{3!} &= \frac{6!}{3!} \\
+&= \frac{720}{6} \\
+&= \boxed{\textbf{(E) }120}
+\end{align*}
+</cmath>
+== See Also ==
+{{AHSME box|year=1996|num-b=2|num-a=4}}
+{{MAA Notice}}
+[[Category:Introductory Algebra Problems]]