Difference between revisions of "2002 AMC 10B Problems/Problem 9"
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== Problem == | == Problem == | ||
| − | Using the letters <math>A</math>, <math>M</math>, <math>O</math>, <math>S</math>, and <math>U</math>, we can form | + | Using the letters <math>A</math>, <math>M</math>, <math>O</math>, <math>S</math>, and <math>U</math>, we can form five-letter "words". If these "words" are arranged in alphabetical order, then the "word" <math>USAMO</math> occupies position |
<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> | ||
| − | == Solution == | + | == 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>. |
| − | == Solution == | + | == 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> |
| + | |||
| + | == Video Solution by Omega Learn == | ||
| + | https://youtu.be/RldWnL4-BfI?t=676 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2002|ab=B|num-b=8|num-a=10}} | ||
| + | |||
| + | [[Category:Introductory Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 03:20, 4 November 2022
Problem
Using the letters 
, 
, 
, 
, and 
, we can form five-letter "words". If these "words" are arranged in alphabetical order, then the "word" 
occupies position
Solution 1
There are 
"words" beginning with each of the first four letters alphabetically. From there, there are 
with as the first letter and each of the first three letters alphabetically. After that, the next "word" is , hence our answer is 
.
Solution 2
Let 
, 
, 
, 
, and 
. Then counting backwards, 
, so the answer is 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
