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AMC10每日一题（2000年真题#13）

2018-01-02
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2000 AMC 10 竞赛试题/第13题

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 (三角形的) 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

In each column there must be one yellow peg. In particular, in the rightmost (最右边的) 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 (逻辑), 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(