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

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== Problem ==
 
== 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 <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 N/<math>10</math>.
  
 
== Solution ==
 
== Solution ==
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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>.
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Thus, <math>N = 2520 + 1200 = 3720</math>, and <math>\frac{N}{10} = \boxed{372}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 21:16, 21 January 2020

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/$10$.

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 $\frac{N}{10} = \boxed{372}$.

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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