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> | ||
| − | == 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>. | 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 == | ||
| + | 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> | ||
==See Also== | ==See Also== | ||
Revision as of 13:42, 15 January 2018
Contents
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 
. Then counting backwards, 
, so the answer is 
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. 
