2002 AMC 10B Problems/Problem 9

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 1

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 $4\cdot 4!+3\cdot 3!+1=\boxed{115\Rightarrow\text{(D)}}$ .

Solution 2

Let $A = 1$ , $M = 2$ , $O = 3$ , $S = 4$ , and $U = 5$ . Then counting backwards, $54321, 54312, 54231, 54213, 54132, 54123$ , so the answer is $\boxed{115\Rightarrow\text{(D)}}$

Video Solution

https://youtu.be/wopflrvUN2c?t=1346

~ pi_is_3.14

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AMC 10 Problems and Solutions

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2002 AMC 10B Problems/Problem 9

Contents

Problem

Solution 1

Solution 2

Video Solution

See Also