Difference between revisions of "2002 AMC 10B Problems/Problem 9"

Revision as of 10:22, 4 July 2013

Problem

Using the letters $A$ , $M$ , $O$ , $S$ , and $U$ , we can form five-letter "words". If these "words" are arranged in alphabetical order, then the "word" $USAMO$ occupies position

$\mathrm{(A) \ } 112\qquad \mathrm{(B) \ } 113\qquad \mathrm{(C) \ } 114\qquad \mathrm{(D) \ } 115\qquad \mathrm{(E) \ } 116$

Solution

There are $4!\cdot 4$ "words" beginning with each of the first four letters alphabetically. From there, there are $3!\cdot 3$ with $U$ as the first letter and each of the first three letters alphabetically. After that, the next "word" is $USAMO$ , hence our answer is $4\cdot 4!+3\cdot 3!+1=\boxed{115\Rightarrow\text{(D)}}$ .

2002 AMC 10B (Problems • Answer Key • Resources)
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Difference between revisions of "2002 AMC 10B Problems/Problem 9"

Revision as of 10:22, 4 July 2013

Problem

Solution

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