Art of Problem Solving

2000 AMC 10 Problems/Problem 13: Difference between revisions

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New page: The question is rather ambiguous, however I will assume that the pegs of the same color are distinguishable. Clearly, there is only 1 possible ordering if the colors are indistinguishable...
 
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==Problem==
==Solution==
The question is rather ambiguous, however I will assume that the pegs of the same color are distinguishable.
The question is rather ambiguous, however I will assume that the pegs of the same color are distinguishable.


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Or, C.
Or, C.
==See Also==
{{AMC10 box|year=2000|num-b=12|num-a=14}}

Revision as of 18:50, 8 January 2009

Problem

Solution

The question is rather ambiguous, however I will assume that the pegs of the same color are distinguishable.

Clearly, there is only 1 possible ordering if the colors are indistinguishable.

Thus, $5! \cdot 4! \cdot 3! \cdot 2! \cdot 1!$

Or, C.

See Also

2000 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions