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

(Solution)
m (Solution 2)
Line 9: Line 9:
  
 
== Solution 2 ==
 
== Solution 2 ==
Let <math>A = 1, \ldots, U = 5</math>. Then counting backwards, <math>54321, 54312, 54231, 54213, 54132, 54123</math>, so the answer is <math>\boxed{115\Rightarrow\text{(D)}}</math>
+
Let <math>A = 1</math>, <math>M = 2</math>, <math>O = 3</math>, <math>S = 4</math>, and <math>U = 5</math>. Then counting backwards, <math>54321, 54312, 54231, 54213, 54132, 54123</math>, so the answer is <math>\boxed{115\Rightarrow\text{(D)}}</math>
  
 
==See Also==
 
==See Also==

Revision as of 21:52, 10 August 2020

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 $\boxed{115\Rightarrow\text{(D)}}$

See Also

2002 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS