Question 338940: The problem is about color arrangements . I solved it but don't know if the answer is right . If someone could review my work I would be very grateful.
Each of six adjacent squares in a strip is to be filled with anyone of ten possible colors. How many ways are there of coloring the strip so that no two adjacent squares have the same color?
in my solution I perceived that there each square beside the first one could be filled with any nine of the ten colors. and that the first square could be filled with any of the ten colors.With six squares that amounted of 9+9+9+9+9+10=55 , 55 possible positions , that would be my n. and 6 would be r. I used the Factorial Form for P(n,r): P(n,r)=n!/(n-r)!
55!/(55-6)!=55!/49!=55*54*53*52*51*50=aprox 2.087*10^10
Did I do it right or am I way off?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Each of six adjacent squares in a strip is to be filled with anyone of ten possible colors. How many ways are there of coloring the strip so that no two adjacent squares have the same color?
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1st square: 10 ways
2nd square: 9 ways
3rd square: 9 ways
4th square: 9 ways
5th square: 9 ways
6th square: 9 ways
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Total # of ways: 10*9^5 = 590490 ways
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Cheers,
Stan H.
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