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

Revision as of 20:11, 18 August 2011

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

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

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 7: / Line 7: @@
 ==Solution==
-The numerator is <math>(3!)! = 6!.
+The numerator is <math>(3!)! = 6!</math>.
-The denominator is </math>3! = 6<math>.
+The denominator is <math>3! = 6</math>.
-Using the property that </math>6! = 6 \cdot 5!<math> in the numerator, the sixes cancel, leaving </math>5! = 120<math>, which is answer </math>\boxed{E}$.
+Using the property that <math>6! = 6 \cdot 5!</math> in the numerator, the sixes cancel, leaving <math>5! = 120</math>, which is answer <math>\boxed{E}</math>.
 ==See also==
 {{AHSME box|year=1996|num-b=2|num-a=4}}