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? | ||
+ | |||
+ | <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> | ||
+ | |||
+ | |||
+ | <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 21:28, 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?
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,
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 |