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Difference between revisions of "2000 AMC 10 Problems/Problem 13"

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==Problem==
 
==Problem==
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There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?
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<asy>
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unitsize(20);
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dot((0,0));
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dot((1,0));
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dot((2,0));
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dot((3,0));
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dot((4,0));
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dot((0,1));
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dot((1,1));
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dot((2,1));
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dot((3,1));
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dot((0,2));
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dot((1,2));
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dot((2,2));
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dot((0,3));
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dot((1,3));
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dot((0,4));
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</asy>
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<math>\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 5!\cdot 4!\cdot 3!\cdot 2!\cdot 1!  \qquad\mathrm{(D)}\ \frac{15!}{5!\cdot 4!\cdot 3!\cdot 2!\cdot 1!} \qquad\mathrm{(E)}\ 15!</math>
  
 
==Solution==
 
==Solution==

Revision as of 22:28, 9 January 2009

Problem

There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?

[asy] unitsize(20); dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((0,3)); dot((1,3)); dot((0,4)); [/asy]


$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 5!\cdot 4!\cdot 3!\cdot 2!\cdot 1! \qquad\mathrm{(D)}\ \frac{15!}{5!\cdot 4!\cdot 3!\cdot 2!\cdot 1!} \qquad\mathrm{(E)}\ 15!$

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 (Problems • Answer Key • Resources)
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
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