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

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<math> \mathrm{(A) \ } 112\qquad \mathrm{(B) \ } 113\qquad \mathrm{(C) \ } 114\qquad \mathrm{(D) \ } 115\qquad \mathrm{(E) \ } 116 </math>
 
<math> \mathrm{(A) \ } 112\qquad \mathrm{(B) \ } 113\qquad \mathrm{(C) \ } 114\qquad \mathrm{(D) \ } 115\qquad \mathrm{(E) \ } 116 </math>
  
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== Solution ==
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== Solution 1 ==
 
There are <math>4!\cdot 4</math> "words" beginning with each of the first four letters alphabetically. From there, there are <math>3!\cdot 3</math> with <math>U</math> as the first letter and each of the first three letters alphabetically. After that, the next "word" is <math>USAMO</math>, hence our answer is <math>4\cdot 4!+3\cdot 3!+1=\boxed{115\Rightarrow\text{(D)}}</math>.
 
There are <math>4!\cdot 4</math> "words" beginning with each of the first four letters alphabetically. From there, there are <math>3!\cdot 3</math> with <math>U</math> as the first letter and each of the first three letters alphabetically. After that, the next "word" is <math>USAMO</math>, hence our answer is <math>4\cdot 4!+3\cdot 3!+1=\boxed{115\Rightarrow\text{(D)}}</math>.
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== Solution 2 ==
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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>
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== Video Solution by Omega Learn ==
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https://youtu.be/RldWnL4-BfI?t=676
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~ pi_is_3.14
  
 
==See Also==
 
==See Also==
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[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 03:20, 4 November 2022

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 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

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