SOLUTION: Let {{{ l_1 }}} and {{{ l_2 }}} two parallel lines. We choose {{{ 8 }}} points on {{{ l_1 }}} and {{{ 7 }}} on {{{ l_2 }}}. Then all the market points on {{{ l_1 }}} are joined wi
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Permutations
-> SOLUTION: Let {{{ l_1 }}} and {{{ l_2 }}} two parallel lines. We choose {{{ 8 }}} points on {{{ l_1 }}} and {{{ 7 }}} on {{{ l_2 }}}. Then all the market points on {{{ l_1 }}} are joined wi
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Question 1171383: Let and two parallel lines. We choose points on and on . Then all the market points on are joined with the market points on . How many points are obtained as an intersection (that are not in nor in ) of the traced segments?
I´ve solved the problem by doing all the lines and I got intersections, so, I´m looking for a solution with out making all of them. Answer by ikleyn(52777) (Show Source):
The problem ASSUMES that the points are in "GENERAL PLACEMENT", which means that among the connecting lines
there are NO parallel lines and that the intersection points are ALL DIFFERENT (there is no coinciding intersection points).
In all, there are 8*7 different lines.
Every two different lines have one intersection point on the plane (inside the strip between the lines or outside it).
The intersection points that lie on the given parallel lines are included in this counting.
Thus the number of intersection points is equal to the number of pairs of different lines,
which is the number of combinations of 56 lines taken 2 at a time = = 1540. ANSWERANSWER. The number of intersection points on the plane is 1540.
The intersection points on the parallel lines are included.
