A flag has 3 horizontal stripes with a vertical bar on either side. With 10 different colors how many different ways are there to color this five region flag with no two touching regions the same. I get $(10\cdot(9\cdot8\cdot9))+(10\cdot9\cdot(8\cdot7\cdot9))=46800$ but I am told the right answer is $41040$. Have I gone wrong?
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2How did you get your result? (Why did you multiply those exact numbers?) That information would be really helpful in order to be able to tell where you made a mistake. – Jan 16 '18 at 01:48
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You can have the following configurations of the flag:
$$\left[\begin{array}{c|c|c} & A & \\ C & B & C \\ & A &\end{array}\right]$$ $$\left[\begin{array}{c|c|c} & A & \\ C & B & D \\ & A &\end{array}\right]$$ $$\left[\begin{array}{c|c|c} & A & \\ D & B & D \\ & C &\end{array}\right]$$ $$\left[\begin{array}{c|c|c} & A & \\ D & B & E \\ & C &\end{array}\right]$$
where $A, B, C, D, E$ are different colours.
For each of those, we need to choose, respectively, $3, 4, 4$ or $5$ colours out of $10$. Thus, the result is $(10\cdot 9\cdot 8)+2(10\cdot 9\cdot 8\cdot 7)+(10\cdot 9\cdot 8\cdot 7\cdot 6)=41040$
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Thanks when I see the answer it seems easy and of course you are right. I thought about it incorrectly. I first assumed both vertical bars the same colour ie 10 possible, then 9 top strip, 8 middle and 9 possible for the bottom strip. Then with 2 colours for the verticals 10x9 times 8 possible for the first horizontal, 7 for the second and 8 for the bottom (mistyped above the last number should be 8). I thought it made sense but clearly I have missed. Thanks again – John Buer Jan 16 '18 at 04:26
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I think you are close to the answer - the 1st case needs to be $10\cdot9\cdot8\cdot 8$ because the last colour can be none of the colour of the vertical bar or middle stripe. Similar for the second - it should be $10\cdot9\cdot 8\cdot 7\cdot 7$. Actually your first addend supposedly accounts for my 1st and 3rd configuration, and the 2nd - for 2nd and 4th. – Jan 16 '18 at 07:39
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