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

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== Problem ==
 
== Problem ==
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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 <math>2007</math>. No character may appear in a [[sequence]] more times than it appears among the four letters in AIME or the four digits in <math>2007</math>. A set of plates in which each possible sequence appears exactly once contains <math>N</math> license plates. Find <math>\frac{N}{10}</math>.
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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 <math>\frac{N}{10}</math>.
  
 
== Solution ==
 
== Solution ==
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*If <math>0</math> appears 0 or 1 times amongst the sequence, there are <math>\frac{7!}{(7-5)!} = 2520</math> sequences possible.
 
*If <math>0</math> appears 0 or 1 times amongst the sequence, there are <math>\frac{7!}{(7-5)!} = 2520</math> sequences possible.
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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>.
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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. In total, that gives us <math>10 \cdot 120 = 1200</math>.
  
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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} = \boxed{372}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 13:01, 3 December 2023

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. In total, that gives us $10 \cdot 120 = 1200$.

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

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

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