Number of triangles formed

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There are 10 points in a plane and 4 of them are collinear.
Find the number of triangles formed by producing the lines resulting from joining the points infinitely in both directions (assuming no two lines are parallel).



I can see that there are 10C2-4C2=40 straight lines. If no two pairs of lines were concurrent we would have 40C3=9880 triangles. However, I do not know how to adjust for the concurrent lines. Any help/suggestions much appreciated.







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    up vote
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    down vote

    favorite












    There are 10 points in a plane and 4 of them are collinear.
    Find the number of triangles formed by producing the lines resulting from joining the points infinitely in both directions (assuming no two lines are parallel).



    I can see that there are 10C2-4C2=40 straight lines. If no two pairs of lines were concurrent we would have 40C3=9880 triangles. However, I do not know how to adjust for the concurrent lines. Any help/suggestions much appreciated.







    share|cite|improve this question






















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      There are 10 points in a plane and 4 of them are collinear.
      Find the number of triangles formed by producing the lines resulting from joining the points infinitely in both directions (assuming no two lines are parallel).



      I can see that there are 10C2-4C2=40 straight lines. If no two pairs of lines were concurrent we would have 40C3=9880 triangles. However, I do not know how to adjust for the concurrent lines. Any help/suggestions much appreciated.







      share|cite|improve this question












      There are 10 points in a plane and 4 of them are collinear.
      Find the number of triangles formed by producing the lines resulting from joining the points infinitely in both directions (assuming no two lines are parallel).



      I can see that there are 10C2-4C2=40 straight lines. If no two pairs of lines were concurrent we would have 40C3=9880 triangles. However, I do not know how to adjust for the concurrent lines. Any help/suggestions much appreciated.









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 29 at 7:39









      TryinHard

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          2 Answers
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          In the statement, "assuming no two lines are parallel" probably means that the
          $$binom102-binom42+1=40$$
          lines are distinct and pairwise concurrent. Then the number of triangles is $binom403=9880$.






          share|cite|improve this answer





























            up vote
            0
            down vote













            Actually, I think I was on the right track. The number of triangles eliminated for each of the 4 collinear points is 7C3=35 and the number of triangles eliminated from the 6 non-collinear points is 9C3=84.
            Hence we eliminate 4*35+6*84=644 triangles, resulting in 9880-644=9236 triangles formed.






            share|cite|improve this answer




















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              1
              down vote













              In the statement, "assuming no two lines are parallel" probably means that the
              $$binom102-binom42+1=40$$
              lines are distinct and pairwise concurrent. Then the number of triangles is $binom403=9880$.






              share|cite|improve this answer


























                up vote
                1
                down vote













                In the statement, "assuming no two lines are parallel" probably means that the
                $$binom102-binom42+1=40$$
                lines are distinct and pairwise concurrent. Then the number of triangles is $binom403=9880$.






                share|cite|improve this answer
























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  In the statement, "assuming no two lines are parallel" probably means that the
                  $$binom102-binom42+1=40$$
                  lines are distinct and pairwise concurrent. Then the number of triangles is $binom403=9880$.






                  share|cite|improve this answer














                  In the statement, "assuming no two lines are parallel" probably means that the
                  $$binom102-binom42+1=40$$
                  lines are distinct and pairwise concurrent. Then the number of triangles is $binom403=9880$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 29 at 8:22

























                  answered Aug 29 at 8:02









                  Robert Z

                  85.5k1055123




                  85.5k1055123




















                      up vote
                      0
                      down vote













                      Actually, I think I was on the right track. The number of triangles eliminated for each of the 4 collinear points is 7C3=35 and the number of triangles eliminated from the 6 non-collinear points is 9C3=84.
                      Hence we eliminate 4*35+6*84=644 triangles, resulting in 9880-644=9236 triangles formed.






                      share|cite|improve this answer
























                        up vote
                        0
                        down vote













                        Actually, I think I was on the right track. The number of triangles eliminated for each of the 4 collinear points is 7C3=35 and the number of triangles eliminated from the 6 non-collinear points is 9C3=84.
                        Hence we eliminate 4*35+6*84=644 triangles, resulting in 9880-644=9236 triangles formed.






                        share|cite|improve this answer






















                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          Actually, I think I was on the right track. The number of triangles eliminated for each of the 4 collinear points is 7C3=35 and the number of triangles eliminated from the 6 non-collinear points is 9C3=84.
                          Hence we eliminate 4*35+6*84=644 triangles, resulting in 9880-644=9236 triangles formed.






                          share|cite|improve this answer












                          Actually, I think I was on the right track. The number of triangles eliminated for each of the 4 collinear points is 7C3=35 and the number of triangles eliminated from the 6 non-collinear points is 9C3=84.
                          Hence we eliminate 4*35+6*84=644 triangles, resulting in 9880-644=9236 triangles formed.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Sep 5 at 10:27









                          TryinHard

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