Difference between revisions of "2007 AIME II Problems/Problem 1"

(Problem)
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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>\displaystyle N = 2520 + 1200 = 3720</math>, and <math>\frac{N}{10} = 372</math>.
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Thus, <math>N = 2520 + 1200 = 3720</math>, and <math>\frac{N}{10} = 372</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
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{{MAA Notice}}

Revision as of 22:32, 4 July 2013

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 $\frac{N}{10}$.

Solution

There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice.

  • If $0$ appears 0 or 1 times amongst the sequence, there are $\frac{7!}{(7-5)!} = 2520$ sequences possible.
  • If $0$ appears twice in the sequence, there are ${5\choose2} = 10$ places to place the $0$s. There are $\frac{6!}{(6-3)!} = 120$ ways to place the remaining three characters. Totally, that gives us $10 \cdot 120 = 1200$.

Thus, $N = 2520 + 1200 = 3720$, and $\frac{N}{10} = 372$.

See also

2007 AIME II (ProblemsAnswer KeyResources)
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

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