I have a $2\times3$ grid which is colored with four colours. How many colourings do there exists such that every two adjacent squares have different colours? I don't know how to start.
Grid $2\times3$ colored with four colours
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combinatorics
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0Start by trying to solve similar problems for smaller grids and fewer colors. – 2017-01-14
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0@MJD What is the basic idea? – 2017-01-14
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1This idea should work but not in general: Top middle cell has 4 choices, neighbors have 3 choices, then take cases for the last two cells. – 2017-01-14
1 Answers
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Color the grid by columns (top and bottom cells) going from left to right:
You have $4\cdot 3=12$ ways to color the $1$st column. Now given that you've made your choice of colors for the $1$st column, the number of ways to color the $2$nd column will drop down from $12$ to $7$. The same happens for the 3rd column, once you've colored the $2$nd one . So the total number is $$12\cdot 7 \cdot 7=588.$$
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0WHy 7? They are 6? – 2017-01-14
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0Let A, B,C, D be the four colors. Suppose the $1$st column is (A,B) (A on the top, and B on the bottom). Can you list the possibilities for the $2$nd column? – 2017-01-14
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0Sorry, you have right! – 2017-01-14