Difference between revisions of "2005 AIME I Problems/Problem 5"

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There are <math>{8\choose4} = 70</math> ways to position the gold coins in the stack of 8 coins, which determines the positions of the silver coins.
 
There are <math>{8\choose4} = 70</math> ways to position the gold coins in the stack of 8 coins, which determines the positions of the silver coins.
  
Create a string of letters H and T to denote the orientation of the top of the coin. To avoid making two faces touch, we cannot have the arrangement HT. Thus, all possible configurations must be a string of tails followed by a string of heads, since after the first H no more tails can appear. The first H can occur in a maximum of eight times different positions, and then there is also the possibility that it doesn’t occur at all, for <math>9</math> total configurations. Thus, the answer is <math>70 \cdot 9 = 630</math>.
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Create a string of letters H and T to denote the orientation of the top of the coin. To avoid making two faces touch, we cannot have the arrangement HT. Thus, all possible configurations must be a string of tails followed by a string of heads, since after the first H no more tails can appear. The first H can occur in a maximum of eight times different positions, and then there is also the possibility that it doesn’t occur at all, for <math>9</math> total configurations. Thus, the answer is <math>70 \cdot 9 = \boxed{630}</math>.
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== See also ==
 
== See also ==
 
{{AIME box|year=2005|n=I|num-b=4|num-a=6}}
 
{{AIME box|year=2005|n=I|num-b=4|num-a=6}}
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]

Revision as of 22:33, 28 December 2011

Problem

Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangements of the 8 coins.

Solution

There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation.

There are ${8\choose4} = 70$ ways to position the gold coins in the stack of 8 coins, which determines the positions of the silver coins.

Create a string of letters H and T to denote the orientation of the top of the coin. To avoid making two faces touch, we cannot have the arrangement HT. Thus, all possible configurations must be a string of tails followed by a string of heads, since after the first H no more tails can appear. The first H can occur in a maximum of eight times different positions, and then there is also the possibility that it doesn’t occur at all, for $9$ total configurations. Thus, the answer is $70 \cdot 9 = \boxed{630}$.

See also

2005 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions