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I was preparing for my Engineering entrance exam and got across this question.

A flag is to be made with 6 vertical stripes by using colours yellow, blue, green and red in such a way that none of the adjacent stripes have same colour. How many ways is it possible?

I landed up with 12x81 but I tried again and landed with 24x216

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  • $\begingroup$ if you explain how you counted we can find possible errors $\endgroup$ – mm-aops Jun 21 '14 at 18:27
  • $\begingroup$ @mm-aops first I took yellow in position 1 then 2 then on till 6 later green, red, blue. So 4 colours and 6 different positions. 6x4=24. There was an option given there 24x216 so i thought it might be right. $\endgroup$ – user3627194 Jun 21 '14 at 18:32
  • $\begingroup$ Just worked on this same type problem over here: math.stackexchange.com/questions/840965/permutation-combination/… $\endgroup$ – bobbym Jun 21 '14 at 18:40
  • $\begingroup$ @bobbym ok. But how did u expand ( 1111), that vertical one in end $\endgroup$ – user3627194 Jun 21 '14 at 18:42
  • $\begingroup$ Hi; I had to use Mathematica, too tedious to do by hand. You will find more manageable solutions there too. $\endgroup$ – bobbym Jun 21 '14 at 18:44
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what you do is you consider the first strip - its colour can be chosen out of all possible $4$. once you chose the colour of the first one you only have $3$ choices for the second one - cause it can't be the same which gives you $4 \cdot 3$ possibilities for the first two stripes. continuing this way you end up with $4 \cdot 3^5$ ways

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  • $\begingroup$ There's no option as such there. $\endgroup$ – user3627194 Jun 21 '14 at 18:36
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    $\begingroup$ I'm pretty sure that's correct unless you have extra assumptions in your problem - i.e. all the colors have to be used $\endgroup$ – mm-aops Jun 21 '14 at 18:41
  • $\begingroup$ I agree that is correct. It also is one of his options of 12 x 81. $\endgroup$ – bobbym Jun 21 '14 at 19:09
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It must be 12 x 81 for sure. It is not that hard. For six strips the first can be filled up in 4 ways, i.e, any one of the colours ; and the next can be filled in 3 ways since the colour used in the first strip cannot be used. The third can be filled up again in 3 ways since the colour to the left cannot be used. And this goes on till the last strip. So we finally get 4 x 3^5. Equivalent to 12 x 81.

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