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>.") |
m (see also box, statement of problem) |
||
(One intermediate revision by one other user not shown) | |||
Line 1: | Line 1: | ||
+ | == Problem == | ||
+ | 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. | ||
+ | |||
==Solution== | ==Solution== | ||
− | + | 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>. | |
+ | |||
+ | == See Also == | ||
+ | {{Mock AIME box|year=2010|n=1|num-b=7|num-a=9}} | ||
+ | [[Category:Intermediate Combinatorics Problems]] |
Latest revision as of 13:38, 2 August 2024
Problem
In the context of this problem, a is a
block, a
is a
block, and a
is a
block. If
is the number of ways George can place one square, two identical dominoes, and three identical triominoes on a
chessboard such that no two overlap, find the remainder when
is divided by 1000.
Solution
Once the pieces are placed, there will be blank spaces. So, we are simply ordering
triominoes,
dominoes,
square, and
blank spaces. That's just
, with last three digits
.
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 |