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

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==Problem==
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<math> \frac{(3!)!}{3!}= </math>
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<math> \text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120 </math>
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==Solution==
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The numerator is <math>(3!)! = 6!.
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The denominator is </math>3! = 6<math>.
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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}$.
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==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=2|num-a=4}}
 
{{AHSME box|year=1996|num-b=2|num-a=4}}

Revision as of 20:11, 18 August 2011

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$ (Error compiling LaTeX. ! Missing $ inserted.)3! = 6$.

Using the property that$ (Error compiling LaTeX. ! Missing $ inserted.)6! = 6 \cdot 5!$in the numerator, the sixes cancel, leaving$5! = 120$, which is answer$\boxed{E}$.

See also

1996 AHSME (ProblemsAnswer KeyResources)
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
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