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

Revision as of 19:32, 26 July 2017

LOTS OF POOP

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

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 == 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>.
+LOTS OF POOP
 == Solution ==

2007 AIME II (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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All AIME Problems and Solutions