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

Revision as of 20:16, 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 10 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 $\frac{N}{10} = \boxed{372}$ .

@@ Line 3: / Line 3: @@
 == Solution ==
-There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice.
+There are 10 different characters that can be picked, with 0 being the only number that can be repeated twice.
 *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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Difference between revisions of "2007 AIME II Problems/Problem 1"

Revision as of 20:16, 21 January 2020

Problem

Solution

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