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

Latest revision as of 13:06, 5 July 2013

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

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

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 $\boxed{E}$ .

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 15: / Line 15: @@
 ==See also==
 {{AHSME box|year=1996|num-b=2|num-a=4}}
+{{MAA Notice}}