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

(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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The question is rather ambiguous, however I will assume that the pegs of the same color are distinguishable.
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==Problem==
  
Clearly, there is only 1 possible ordering if the colors are indistinguishable.
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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?
  
Thus, <math>5! \cdot 4! \cdot 3! \cdot 2! \cdot 1!</math>
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<asy>
 +
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));
 +
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>
  
Or, C.
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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>
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==Videos:==
 +
https://www.youtube.com/watch?v=IP3aip6lT40
 +
 
 +
==Solution==
 +
 
 +
In each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there.
 +
 
 +
In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left for the yellow peg in this column.
 +
 
 +
By similar logic, we can fill in the yellow pegs as shown:
 +
 
 +
<asy>
 +
unitsize(20);
 +
dot((0,0));
 +
dot((1,0));
 +
dot((2,0));
 +
dot((3,0));
 +
label("Y",(4,-.35),N);
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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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label("Y",(3,.6),N);
 +
dot((0,2));
 +
dot((1,2));
 +
label("Y",(2,1.6),N);
 +
dot((0,3));
 +
label("Y",(1,2.6),N);
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label("Y",(0,3.6),N);
 +
</asy>
 +
 
 +
After this we can proceed to fill in the whole pegboard, so there is only <math>1</math> arrangement of the pegs. The answer is <math>\boxed{\text{B}}</math>
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 +
<asy>
 +
unitsize(20);
 +
label("O",(0,-.35),N);
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label("B",(1,-.35),N);
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label("G",(2,-.35),N);
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label("R",(3,-.35),N);
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label("Y",(4,-.35),N);
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label("B",(0,.6),N);
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label("G",(1,.6),N);
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label("R",(2,.6),N);
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label("Y",(3,.6),N);
 +
label("G",(0,1.6),N);
 +
label("R",(1,1.6),N);
 +
label("Y",(2,1.6),N);
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label("R",(0,2.6),N);
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label("Y",(1,2.6),N);
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label("Y",(0,3.6),N);
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</asy>
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 +
==See Also==
 +
 
 +
{{AMC10 box|year=2000|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Latest revision as of 17:04, 19 October 2020

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]


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

Videos:

https://www.youtube.com/watch?v=IP3aip6lT40

Solution

In each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there.

In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left for the yellow peg in this column.

By similar logic, we can fill in the yellow pegs as shown:

[asy] unitsize(20); dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); label("Y",(4,-.35),N); dot((0,1)); dot((1,1)); dot((2,1)); label("Y",(3,.6),N); dot((0,2)); dot((1,2)); label("Y",(2,1.6),N); dot((0,3)); label("Y",(1,2.6),N); label("Y",(0,3.6),N); [/asy]

After this we can proceed to fill in the whole pegboard, so there is only $1$ arrangement of the pegs. The answer is $\boxed{\text{B}}$

[asy] unitsize(20); label("O",(0,-.35),N); label("B",(1,-.35),N); label("G",(2,-.35),N); label("R",(3,-.35),N); label("Y",(4,-.35),N); label("B",(0,.6),N); label("G",(1,.6),N); label("R",(2,.6),N); label("Y",(3,.6),N); label("G",(0,1.6),N); label("R",(1,1.6),N); label("Y",(2,1.6),N); label("R",(0,2.6),N); label("Y",(1,2.6),N); label("Y",(0,3.6),N); [/asy]

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

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