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!</math>.
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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}</math>.
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==See also==
 
==See also==
{{AHSME box|year=1995|num-b=7|num-a=9}}
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{{AHSME box|year=1996|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 14:06, 5 July 2013

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 $\boxed{E}$.

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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