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Difference between revisions of "Mock AIME 1 2010 Problems/Problem 8"

(Created page with "==Solution== This is the same thing as arranging 1 A, 2 Bs, and 3 Cs to form a word. Our answer is <math>\frac{6!}{3!2!1!}=\boxed{60}</math>.")
 
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== Problem ==
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In the context of this problem, a <math>\emph{square}</math> is a <math>1 \times 1</math> block, a <math>\emph{domino}</math> is a <math>1 \times 2</math> block, and a <math>\emph{triomino}</math> is a <math>1 \times 3</math> block. If <math>N</math> is the number of ways George can place one square, two identical dominoes, and three identical triominoes on a <math>1 \times 20</math> chessboard such that no two overlap, find the remainder when <math>N</math> is divided by 1000.
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==Solution==
 
==Solution==
  
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This is the same thing as arranging 1 A, 2 Bs, and 3 Cs to form a word. Our answer is <math>\frac{6!}{3!2!1!}=\boxed{60}</math>.
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Once the pieces are placed, there will be <math>20-3(3)-2(2)-1=6</math> blank spaces. So, we are simply ordering <math>3</math> triominoes, <math>2</math> dominoes, <math>1</math> square, and <math>6</math> blank spaces. That's just <math>\frac{12!}{3!2!6!} = 55440</math>, with last three digits <math>\boxed{440}</math>.
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== See Also ==
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{{Mock AIME box|year=2010|n=1|num-b=7|num-a=9}}
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[[Category:Intermediate Combinatorics Problems]]

Latest revision as of 12:38, 2 August 2024

Problem

In the context of this problem, a $\emph{square}$ is a $1 \times 1$ block, a $\emph{domino}$ is a $1 \times 2$ block, and a $\emph{triomino}$ is a $1 \times 3$ block. If $N$ is the number of ways George can place one square, two identical dominoes, and three identical triominoes on a $1 \times 20$ chessboard such that no two overlap, find the remainder when $N$ is divided by 1000.

Solution

Once the pieces are placed, there will be $20-3(3)-2(2)-1=6$ blank spaces. So, we are simply ordering $3$ triominoes, $2$ dominoes, $1$ square, and $6$ blank spaces. That's just $\frac{12!}{3!2!6!} = 55440$, with last three digits $\boxed{440}$.

See Also

Mock AIME 1 2010 (Problems, Source)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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