Question 475863: There are 10 points in a plane. No three of these points are in a straight line, except 4 points which are all in the same straight line. How many straight lines can be formed by joining the 10 points?
Here is how I did it:
n=10 = points total
r=2 = points required to make a line
10C2 = 45 lines
n=4 = colinear points
r=2 = points required to make a line
4C2= 6 lines made from colinear points
n=6 = points that are not colinear
r=2 = points required to make a line
6C2 = 16 lines made from points which are not colinear
But I am not sure what to do with this now. If I multiply 4C2 with 6C2 I get 15*6 = 90 lines. It does not make sense to multiply. Let's do 15 plus 6 that gives 21 lines are made from 10 points but that also sounds wrong; it's too small.
Can I approach this problem this way?
case 1
n=4 colinear points
each colinear point can make 6 lines; thus 1 linear point/6 lines, thus 4*6 = 24 lines
case 2
n=6 (other points, not alligned)
each not alligned point can make 9 lines, thus 1 non-alligned point/9 lines, thus 6*9 = 54 lines
case 1 + case 2 gives 54+24 = 78 lines can be formed by joining 4 colinear points and 6 not alligned points. But this is not the right answer either.
Would you explain please how to proceed with this problem? Thank you so much for your help.
Yours,
I.
Found 2 solutions by scott8148, richard1234:
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! you are on the right track
case 1 is good
case 2 ___ some of the lines in 6*9 are already accounted for in case 1
___ the 6C2 that you had earlier works good here, lines from non-colinear points
don't forget the ONE colinear line
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! I think the simplest way is to start with the four collinear points, in which we have one line.
The next case would be involving the other six points: choose two of these six points, 6C2 = 15 ways to do it.
The final case would be to choose one of the four collinear points, then one of the six remaining points, in which 24 lines are determined (we do not overcount lines because no three of the other points are collinear).
The number of lines is 1+15+24 = 40.
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