Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Video Solution)
(→‎Solution 2)
 
Line 9: Line 9:
  
 
== Solution 2 ==
 
== Solution 2 ==
−
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>
+
Let <math>A = 1</math>, <math>M = 2</math>, <math>O = 3</math>, <math>S = 4</math>, and <math>U = 5</math>. The ways the numbers can be arranged is <math>5</math> factorial which is <math>120</math>, with <math>12345</math> in spot <math>1</math> and <math>54321</math> in spot <math>120</math>. Then counting backwards, <math>54321, 54312, 54231, 54213, 54132, 54123</math>, so the answer is <math>\boxed{115\Rightarrow\text{(D)}}</math>
  
 
== Video Solution by Omega Learn ==
 
== Video Solution by Omega Learn ==

Latest revision as of 13:56, 28 June 2025

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 $U = 5$. The ways the numbers can be arranged is $5$ factorial which is $120$, with $12345$ in spot $1$ and $54321$ in spot $120$. Then counting backwards, $54321, 54312, 54231, 54213, 54132, 54123$, so the answer is $\boxed{115\Rightarrow\text{(D)}}$

Video Solution by Omega Learn

https://youtu.be/RldWnL4-BfI?t=676

~ pi_is_3.14

See Also

2002 AMC 10B (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems/Problem_9&oldid=252660"