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Math Help - Number of possible triangles

  1. #1
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    Number of possible triangles

    Let S(10)={1,2,3,4,5,6,7,8,9,10}
    and T(10)={ number of triangles formed by using any three elements of S(10) }
    in the same way
    S(20) = {1,2,3,4...,18,19,20}
    and T(20) ={number of triangles formed by using any three elements of S(20)}
    then find the value of T(20) - T(10)
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    Quote Originally Posted by jashansinghal View Post
    Let S(10)={1,2,3,4,5,6,7,8,9,10}
    and T(10)={ number of triangles formed by using any three elements of S(10) }
    in the same way
    S(20) = {1,2,3,4...,18,19,20}
    and T(20) ={number of triangles formed by using any three elements of S(20)}
    then find the value of T(20) - T(10)
    Well you don't define all terms, but I'm assuming that some valid triangles for S(10) are

    Code:
    3
    Code:
    1
    2 3
    4 5 6
    Code:
    6
    1 2
    Code:
    6
    2 1
    Where the last two are considered distinct.

    With notation C(n,k) for binomial coefficient (that is, n choose k) and P(n,k) the number of ordered k-subsets of {1,...,n}; thus P(n,k) = C(n,k)*k!

    Then the number of triangles from S(n) is T(n)=P(n,1)+P(n,3)+P(n,6)+... not letting k exceed n in P(n,k), and where the sequence of k is the triangle numbers 1,3,6,10,15,...
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    IF no sides are the same, then the 1 cannot be used to form a triangle.

    2,3,....,9 : 50 triangles
    2,3,...,19 : 525 triangles
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  4. #4
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    Oh sorry I misread the problem, the part about using three elements of the set to make a triangle. I would go with T(20) - T(10) = P(20,3) - P(10,3).
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    Quote Originally Posted by undefined View Post
    Oh sorry I misread the problem, the part about using three elements of the set to make a triangle. I would go with T(20) - T(10) = P(20,3) - P(10,3).
    No, you can't do that: triangles side lengths have rules: as example, you can't have 2,4,9 ... OK?
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    Quote Originally Posted by Wilmer View Post
    No, you can't do that: triangles side lengths have rules: as example, you can't have 2,4,9 ... OK?
    This is why the OP should have defined terms. If you look at my first post you will see that I did not interpret the elements as side lengths, but rather as arrangements in a triangular fashion, like is commonly done in recreational mathematics.

    I know what the triangle inequality is.. don't patronise me.
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    Is the formula for  T(n) ~~ \frac{1}{4} \left( 2 \binom{n}{3} - \binom{n}{2} + [\frac{n}{2}] \right)    ?
    Last edited by simplependulum; June 18th 2010 at 08:36 PM.
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    Quote Originally Posted by undefined View Post
    ..... but rather as arrangements in a triangular fashion, like is commonly done in recreational mathematics.
    I see; didn't realise that...
    Plus was not "patronizing"; just "typing quickly" !
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    Quote Originally Posted by Wilmer View Post
    I see; didn't realise that...
    Plus was not "patronizing"; just "typing quickly" !
    Haha no problem!
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    Quote Originally Posted by simplependulum View Post
    Is the formula for  T(n) ~~ \frac{1}{4} \left( 2 \binom{n}{3} - \binom{n}{2} + [\frac{n}{2}] \right)    ?
    Well, I don't know...

    IF OP means geometric triangles, then there are 50 in (1,2,.....,10):
    1: 2,3,4
    2: 2,4,5
    ...
    49: 7,9,10
    50: 8,9,10
    Of course, when written in ascending order, sum of 1st two sides must exceed 3rd.
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    Quote Originally Posted by simplependulum View Post
    Is the formula for  T(n) ~~ \frac{1}{4} \left( 2 \binom{n}{3} - \binom{n}{2} + [\frac{n}{2}] \right)    ?
    The formula seems good; see A002623. I'm not sure whether I should ask where you got the formula, or try to find out on my own.
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    write the three sides in ascending order , ie x<y<z , then consider the value of z .

    Then i plot a graph with lattices , 1 <= x <z , 1<= y <z , x < y , x+y > z , the number of lattices satisfying the constraints should be ( just like handling some problems related to linear programming ) :

     \frac{1}{2} \left[ (z-1)^2 - \frac{ z(z-1)}{2}  - \left( z-1 - [ \frac{z}{2} ] \right) \right]

     = \frac{1}{2} \left[ \frac{ (z-1)(z-2) }{2} - (z-1) + [\frac{z}{2}] \right]

    Then take sum from  z =1 to  z = n ....

     \sum \frac{ (z-1)(z-2) }{2} = \binom{n}{3}  ~,~ \sum (z-1) = \binom{n}{2}

    I also find that  \sum [\frac{z}{2}] = \frac{1}{2} \binom{n}{2} + \frac{1}{2} [\frac{n}{2}]

    so  T(n) = \frac{1}{4}\left( 2\binom{n}{3} - \binom{n}{2} + [\frac{n}{2}] \right)
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  13. #13
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    Quote Originally Posted by simplependulum View Post
    write the three sides in ascending order , ie x<y<z , then consider the value of z .

    Then i plot a graph with lattices , 1 <= x <z , 1<= y <z , x < y , x+y > z , the number of lattices satisfying the constraints should be ( just like handling some problems related to linear programming ) :

    ...

    so  T(n) = \frac{1}{4}\left( 2\binom{n}{3} - \binom{n}{2} + [\frac{n}{2}] \right)
    Thanks, definitely wouldn't have come up with that on my own. (Forum doesn't let me give reputation though; the thanks system seemed better as far providing a way for shallow people like me to mindlessly accumulate stats... also a way to show thanks without cluttering the thread with more posts.)
    Last edited by undefined; June 19th 2010 at 01:32 AM.
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    Quote Originally Posted by undefined View Post
    The formula seems good; see A002623.
    Yes. However, A002623 includes triangles like 1-1-1 and 1-2-2...
    which I guess is why if n=1, answer is 1: 1-1-1

    A002623: 1, 3, 7, 13, 22, 34, 50, 70, 95, 125, 161, 203, 252, 308, 372, 444, 525, 615, 715, .........
    can then be shown as:
    0, 0, 0, 1, 3, 7, 13, 22, 34, 50**, 70, 95, 125, 161, 203, 252, 308, 372, 444, 525**, 615, 715, ........
    which then would be correct if elements cannot be repeated.
    50** and 525** then is for (1 to 10) and (1 to 20) or n=10 and n=20.
    Last edited by Wilmer; June 19th 2010 at 06:06 AM. Reason: none
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  15. #15
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    Quote Originally Posted by simplependulum View Post
     T(n) ~~ \frac{1}{4} \left( 2 \binom{n}{3} - \binom{n}{2} + [\frac{n}{2}] \right)    ?
    Guys,
    I didnt get this formula and how to use it. Secondly , the series which you gave , I cannot make head or tail out of it. Can you tell me this thing in a simpler way because I need to solve this question urgently. Please help guys.
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