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 == | ||
− | {{ | + | There are <math>4!\cdot 4</math> "words" beginning with each of the first four letters alphabetically. From there, there are 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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+ | ==See Also== | ||
+ | {{AMC10 box|year=2002|ab=B|num-b=8|num-a=10}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] |
Revision as of 13:08, 27 December 2008
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
There are "words" beginning with each of the first four letters alphabetically. From there, there are are with as the first letter and each of the first three letters alphabetically. After that, the next "word" is , hence our 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 |