Difference between revisions of "2007 AIME II Problems/Problem 1"
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*If <math>0</math> appears twice in the sequence, there are <math>{5\choose2} = 10</math> places to place the <math>0</math>s. There are <math>\frac{6!}{(6-3)!} = 120</math> ways to place the remaining three characters. Totally, that gives us <math>10 \cdot 120 = 1200</math>. | *If <math>0</math> appears twice in the sequence, there are <math>{5\choose2} = 10</math> places to place the <math>0</math>s. There are <math>\frac{6!}{(6-3)!} = 120</math> ways to place the remaining three characters. Totally, that gives us <math>10 \cdot 120 = 1200</math>. | ||
| − | Thus, <math>N = 2520 + 1200 = 3720</math>, and <math>\frac{N}{10} = 372</math>. | + | Thus, <math>N = 2520 + 1200 = 3720</math>, and <math>\frac{N}{10} = \boxed{372}</math>. |
== See also == | == See also == | ||
Revision as of 13:51, 11 August 2018
Problem
A mathematical organization is producing a set of commemorative license plates. Each plate contains a sequence of five characters chosen from the four letters in AIME and the four digits in 2007. No character may appear in a sequence more times than it appears among the four letters in AIME or the four digits in 2007. A set of plates in which each possible sequence appears exactly once contains N license plates. Find N/
.
Solution
There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice.
- If
appears 0 or 1 times amongst the sequence, there are 
sequences possible. - If appears twice in the sequence, there are
places to place the s. There are 
ways to place the remaining three characters. Totally, that gives us 
.
appears 0 or 1 times amongst the sequence, there are 
sequences possible.
places to place the 
ways to place the remaining three characters. Totally, that gives us 
.Thus, 
, and 
.
See also
| 2007 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
