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Math Help - Number of Traingle from octagon

  1. #1
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    Number of Traingle from octagon

    The number triangle whose vertices are at the vertices of an octagon,but none of the whose sides happen to come from the sides of the octagon.

    what i have tried is,

    Number of ways to choose three vertices from octagon is 8C3, and drawing triangle with any of this combination is possible since no three vertices fall on same line.

    To avoid having the octagon edge, we must avoid choosing adjacent vertices. this is equivalent to choosing an edge , or choosing two adjacent edges from octagon to draw triangle.

    Number of ways to choose a single edge is 8 and number of ways to choose two adjacent edges is 7(i.e this by drawing octagon and manually checking. I am not sure how to arrive at this by formula).

    So the answer would be 56 - 7 - 8

    I am totally confused , when doing that edge condition.

    But i am no where near the answer.

    The book says 16 triangles are possible.

    Can anyone help me.
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  2. #2
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    The answer is 16.
    Number the vertices of the octagon, \{1,2,3,4,5,6,7,8\}, consecutively so that 1~\&~8 are adjacent as well j~\&~j+1 for j=1,\cdots,7.
    There are 20 subsets of three which contain no two consecutive integers.
    But those 20 also contain a set like \{1,4,8\}. We donít want that because \overline{1~8} is an edge of the octagon.
    So, how many such subsets must we remove from the 20?
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  3. #3
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    Thanks for your explanation.

    It would be helpful, if you could explain about the 20 subsets.
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  4. #4
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    Quote Originally Posted by kumaran5555 View Post
    Thanks for your explanation.
    It would be helpful, if you could explain about the 20 subsets.
    You should understand that this is not a tutorial service.
    Therefore, you need to have studied the basics.

    How many ways can we arrange the string 11100000 so that no two 1' are next to one another?
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  5. #5
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    I am really sorry for asking such things.

    This is what i have found for this question.

    Using "stars and bars" theorem, we have total of 6 gaps between 0' , 4 in middle and one at each end.

    6C3 would fill the gaps with the necessary condition.

    Please correct my answer if wrong.
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  6. #6
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    Yes that is correct.
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  7. #7
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    I am really sorry to bother you.

    But still i don't get that 20 subsets.

    Please give me a hint at least.

    I am really lost ..
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