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2000 AMC 10 Problem 13

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2000 AMC 10 Problem 13AMC 10 Step by Step
Name:                             Date: September 13, 2026
Arrangements with Restrictions · difficulty 3 of 5 · about 4 min · position P11-15

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?

figure

  1. A)

    00

  2. B)

    11

  3. C)

    5!⋅4!⋅3!⋅2!⋅1!5!\cdot 4!\cdot 3!\cdot 2!\cdot 1!

  4. D)

    15!5!⋅4!⋅3!⋅2!⋅1!\frac{15!}{5!\cdot 4!\cdot 3!\cdot 2!\cdot 1!}

  5. E)

    15!15!

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