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Hello everyone.
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So it is given that using the theorem we have to define the number of non -equivalent colourings of the corners of a triangle that is isisless.
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First, isiseless but not equilateral.
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First with colours red and blue.
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Second with p colours.
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Right.
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So let us first is solve for two colours.
00:24
Right.
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Specifically it is given red and blue.
00:33
Although the colours are of knowing.
00:36
Importance to us just the thing that we have two colors is what is important to us so this can be divided into two case case one which is two vertices are of same color two vertices of same color so this can be done in we can choose from three we can select two in three c2 ways multiplied by now we have two options for color either red or blue so multiplied by two and remaining is 1 vertex which has one color in particular so 1 so this is equal to 3 c2 is equal to 3 multiplied by 2 it becomes 6 now case 2 all are of same color all 3 vertices are of same color so we have total of 2 possibilities either red or do.
01:54
So total becomes 6 plus 2 which is 8.
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Now let us solve for p colors.
02:06
Right.
02:08
So we can make three cases.
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Case 1 all are different.
02:19
So the three vertices can take.
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First can take any p colors.
02:25
Now already we have taken 1 so we are left with p minus 1 colors...