If $12$ distinct points are placed on a circle and all the chords connecting these points are drawn, at how...












1














If $12$ distinct points are placed on the circumference of a circle and all the chords connecting these points are drawn, at how many points do the chords intersect? Assume that no three chords intersect at the same point.



A) $12choose2$



B) $12choose4$



C) $2^{12}$



D) $frac{12!}2$



I tried drawing a circle and tried to find a pattern but couldn't succeed.
for 12 points I found the answer to be $(1+2+....+9)+(1+2+3+.....+8)+....+(1)$ and the result multiplied by $2$. But I'm getting $296$ which is in none of the options. Can anyone help?






combinatorics geometry combinations







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  • Try it for fewer points than $12$ and see if you can spot a pattern.
    – saulspatz
    Dec 12 '18 at 15:08










  • The numbers in (A)-(D) are all usually associated with counting certain (possibly ordered) subsets of a 12-element set. Can you tell what type of subsets these numbers count? Which type of subset corresponds to a single intersection point?
    – Mees de Vries
    Dec 12 '18 at 15:10










  • @saulspatz for 12 points I found the answer to be (1+2+....+9)+(1+2+3+.....+8)+....+(1) and the result multiplied by 2. But I'm getting 296 which is in none of the options
    – Ayaz S Imran
    Dec 12 '18 at 15:14










  • @MeesdeVries (A) is the number of intersection of 12 points
    – Ayaz S Imran
    Dec 12 '18 at 15:16










  • You should add your result, and the method you used to obtain it, to the body of the question. (Don't make another comment. Edit the question.) Then we'll be able to tell you where you've gone wrong.
    – saulspatz
    Dec 12 '18 at 15:16
















1














If $12$ distinct points are placed on the circumference of a circle and all the chords connecting these points are drawn, at how many points do the chords intersect? Assume that no three chords intersect at the same point.



A) $12choose2$



B) $12choose4$



C) $2^{12}$



D) $frac{12!}2$



I tried drawing a circle and tried to find a pattern but couldn't succeed.
for 12 points I found the answer to be $(1+2+....+9)+(1+2+3+.....+8)+....+(1)$ and the result multiplied by $2$. But I'm getting $296$ which is in none of the options. Can anyone help?






combinatorics geometry combinations







share|cite|improve this question
























  • Try it for fewer points than $12$ and see if you can spot a pattern.
    – saulspatz
    Dec 12 '18 at 15:08










  • The numbers in (A)-(D) are all usually associated with counting certain (possibly ordered) subsets of a 12-element set. Can you tell what type of subsets these numbers count? Which type of subset corresponds to a single intersection point?
    – Mees de Vries
    Dec 12 '18 at 15:10










  • @saulspatz for 12 points I found the answer to be (1+2+....+9)+(1+2+3+.....+8)+....+(1) and the result multiplied by 2. But I'm getting 296 which is in none of the options
    – Ayaz S Imran
    Dec 12 '18 at 15:14










  • @MeesdeVries (A) is the number of intersection of 12 points
    – Ayaz S Imran
    Dec 12 '18 at 15:16










  • You should add your result, and the method you used to obtain it, to the body of the question. (Don't make another comment. Edit the question.) Then we'll be able to tell you where you've gone wrong.
    – saulspatz
    Dec 12 '18 at 15:16














1












1








1


0





If $12$ distinct points are placed on the circumference of a circle and all the chords connecting these points are drawn, at how many points do the chords intersect? Assume that no three chords intersect at the same point.



A) $12choose2$



B) $12choose4$



C) $2^{12}$



D) $frac{12!}2$



I tried drawing a circle and tried to find a pattern but couldn't succeed.
for 12 points I found the answer to be $(1+2+....+9)+(1+2+3+.....+8)+....+(1)$ and the result multiplied by $2$. But I'm getting $296$ which is in none of the options. Can anyone help?






combinatorics geometry combinations







share|cite|improve this question















If $12$ distinct points are placed on the circumference of a circle and all the chords connecting these points are drawn, at how many points do the chords intersect? Assume that no three chords intersect at the same point.



A) $12choose2$



B) $12choose4$



C) $2^{12}$



D) $frac{12!}2$



I tried drawing a circle and tried to find a pattern but couldn't succeed.
for 12 points I found the answer to be $(1+2+....+9)+(1+2+3+.....+8)+....+(1)$ and the result multiplied by $2$. But I'm getting $296$ which is in none of the options. Can anyone help?






combinatorics geometry combinations




combinatorics geometry combinations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 16:38









N. F. Taussig

43.6k93355




43.6k93355










asked Dec 12 '18 at 15:04









Ayaz S ImranAyaz S Imran

285




285












  • Try it for fewer points than $12$ and see if you can spot a pattern.
    – saulspatz
    Dec 12 '18 at 15:08










  • The numbers in (A)-(D) are all usually associated with counting certain (possibly ordered) subsets of a 12-element set. Can you tell what type of subsets these numbers count? Which type of subset corresponds to a single intersection point?
    – Mees de Vries
    Dec 12 '18 at 15:10










  • @saulspatz for 12 points I found the answer to be (1+2+....+9)+(1+2+3+.....+8)+....+(1) and the result multiplied by 2. But I'm getting 296 which is in none of the options
    – Ayaz S Imran
    Dec 12 '18 at 15:14










  • @MeesdeVries (A) is the number of intersection of 12 points
    – Ayaz S Imran
    Dec 12 '18 at 15:16










  • You should add your result, and the method you used to obtain it, to the body of the question. (Don't make another comment. Edit the question.) Then we'll be able to tell you where you've gone wrong.
    – saulspatz
    Dec 12 '18 at 15:16


















  • Try it for fewer points than $12$ and see if you can spot a pattern.
    – saulspatz
    Dec 12 '18 at 15:08










  • The numbers in (A)-(D) are all usually associated with counting certain (possibly ordered) subsets of a 12-element set. Can you tell what type of subsets these numbers count? Which type of subset corresponds to a single intersection point?
    – Mees de Vries
    Dec 12 '18 at 15:10










  • @saulspatz for 12 points I found the answer to be (1+2+....+9)+(1+2+3+.....+8)+....+(1) and the result multiplied by 2. But I'm getting 296 which is in none of the options
    – Ayaz S Imran
    Dec 12 '18 at 15:14










  • @MeesdeVries (A) is the number of intersection of 12 points
    – Ayaz S Imran
    Dec 12 '18 at 15:16










  • You should add your result, and the method you used to obtain it, to the body of the question. (Don't make another comment. Edit the question.) Then we'll be able to tell you where you've gone wrong.
    – saulspatz
    Dec 12 '18 at 15:16
















Try it for fewer points than $12$ and see if you can spot a pattern.
– saulspatz
Dec 12 '18 at 15:08




Try it for fewer points than $12$ and see if you can spot a pattern.
– saulspatz
Dec 12 '18 at 15:08












The numbers in (A)-(D) are all usually associated with counting certain (possibly ordered) subsets of a 12-element set. Can you tell what type of subsets these numbers count? Which type of subset corresponds to a single intersection point?
– Mees de Vries
Dec 12 '18 at 15:10




The numbers in (A)-(D) are all usually associated with counting certain (possibly ordered) subsets of a 12-element set. Can you tell what type of subsets these numbers count? Which type of subset corresponds to a single intersection point?
– Mees de Vries
Dec 12 '18 at 15:10












@saulspatz for 12 points I found the answer to be (1+2+....+9)+(1+2+3+.....+8)+....+(1) and the result multiplied by 2. But I'm getting 296 which is in none of the options
– Ayaz S Imran
Dec 12 '18 at 15:14




@saulspatz for 12 points I found the answer to be (1+2+....+9)+(1+2+3+.....+8)+....+(1) and the result multiplied by 2. But I'm getting 296 which is in none of the options
– Ayaz S Imran
Dec 12 '18 at 15:14












@MeesdeVries (A) is the number of intersection of 12 points
– Ayaz S Imran
Dec 12 '18 at 15:16




@MeesdeVries (A) is the number of intersection of 12 points
– Ayaz S Imran
Dec 12 '18 at 15:16












You should add your result, and the method you used to obtain it, to the body of the question. (Don't make another comment. Edit the question.) Then we'll be able to tell you where you've gone wrong.
– saulspatz
Dec 12 '18 at 15:16




You should add your result, and the method you used to obtain it, to the body of the question. (Don't make another comment. Edit the question.) Then we'll be able to tell you where you've gone wrong.
– saulspatz
Dec 12 '18 at 15:16










2 Answers
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If you select any $4$ distinct points on the circle, you'd have one distinct point of intersection. This'll give you a nice little formula of selecting $4$ points out of $n$.



$$N={nchoose4}={12choose4}=495$$






share|cite|improve this answer





























    1














    Let's follow the suggestion of @saulspatz by considering the problem for fewer points. Consider the diagram below.



    chord_intersections_in_a_circle



    The points on each circle have been chosen in such a way that no three chords intersect at the same point. Under these conditions, we can see by inspection that



    begin{array}{c c}
    text{number of points} & text{number of intersections}\ hline
    4 & 1\
    5 & 5\
    6 & 15
    end{array}



    This should suggest a formula for the number of intersections when we have $n$ points and no three chords intersect in the same point.




    A chord is determined by two points of the circle. Two intersecting chords are determined by four points of the circle since the only way the chords can intersect is if we connect both pairs of nonadjacent points.







    share|cite|improve this answer





















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      2 Answers
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      2 Answers
      2






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      active

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      1














      If you select any $4$ distinct points on the circle, you'd have one distinct point of intersection. This'll give you a nice little formula of selecting $4$ points out of $n$.



      $$N={nchoose4}={12choose4}=495$$






      share|cite|improve this answer


























        1














        If you select any $4$ distinct points on the circle, you'd have one distinct point of intersection. This'll give you a nice little formula of selecting $4$ points out of $n$.



        $$N={nchoose4}={12choose4}=495$$






        share|cite|improve this answer
























          1












          1








          1






          If you select any $4$ distinct points on the circle, you'd have one distinct point of intersection. This'll give you a nice little formula of selecting $4$ points out of $n$.



          $$N={nchoose4}={12choose4}=495$$






          share|cite|improve this answer












          If you select any $4$ distinct points on the circle, you'd have one distinct point of intersection. This'll give you a nice little formula of selecting $4$ points out of $n$.



          $$N={nchoose4}={12choose4}=495$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 18:11









          Sauhard SharmaSauhard Sharma

          799117




          799117























              1














              Let's follow the suggestion of @saulspatz by considering the problem for fewer points. Consider the diagram below.



              chord_intersections_in_a_circle



              The points on each circle have been chosen in such a way that no three chords intersect at the same point. Under these conditions, we can see by inspection that



              begin{array}{c c}
              text{number of points} & text{number of intersections}\ hline
              4 & 1\
              5 & 5\
              6 & 15
              end{array}



              This should suggest a formula for the number of intersections when we have $n$ points and no three chords intersect in the same point.




              A chord is determined by two points of the circle. Two intersecting chords are determined by four points of the circle since the only way the chords can intersect is if we connect both pairs of nonadjacent points.







              share|cite|improve this answer


























                1














                Let's follow the suggestion of @saulspatz by considering the problem for fewer points. Consider the diagram below.



                chord_intersections_in_a_circle



                The points on each circle have been chosen in such a way that no three chords intersect at the same point. Under these conditions, we can see by inspection that



                begin{array}{c c}
                text{number of points} & text{number of intersections}\ hline
                4 & 1\
                5 & 5\
                6 & 15
                end{array}



                This should suggest a formula for the number of intersections when we have $n$ points and no three chords intersect in the same point.




                A chord is determined by two points of the circle. Two intersecting chords are determined by four points of the circle since the only way the chords can intersect is if we connect both pairs of nonadjacent points.







                share|cite|improve this answer
























                  1












                  1








                  1






                  Let's follow the suggestion of @saulspatz by considering the problem for fewer points. Consider the diagram below.



                  chord_intersections_in_a_circle



                  The points on each circle have been chosen in such a way that no three chords intersect at the same point. Under these conditions, we can see by inspection that



                  begin{array}{c c}
                  text{number of points} & text{number of intersections}\ hline
                  4 & 1\
                  5 & 5\
                  6 & 15
                  end{array}



                  This should suggest a formula for the number of intersections when we have $n$ points and no three chords intersect in the same point.




                  A chord is determined by two points of the circle. Two intersecting chords are determined by four points of the circle since the only way the chords can intersect is if we connect both pairs of nonadjacent points.







                  share|cite|improve this answer












                  Let's follow the suggestion of @saulspatz by considering the problem for fewer points. Consider the diagram below.



                  chord_intersections_in_a_circle



                  The points on each circle have been chosen in such a way that no three chords intersect at the same point. Under these conditions, we can see by inspection that



                  begin{array}{c c}
                  text{number of points} & text{number of intersections}\ hline
                  4 & 1\
                  5 & 5\
                  6 & 15
                  end{array}



                  This should suggest a formula for the number of intersections when we have $n$ points and no three chords intersect in the same point.




                  A chord is determined by two points of the circle. Two intersecting chords are determined by four points of the circle since the only way the chords can intersect is if we connect both pairs of nonadjacent points.








                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 16:34









                  N. F. TaussigN. F. Taussig

                  43.6k93355




                  43.6k93355






























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