There are 10 different points such that no three of them don't lie on straight line. This means if you select any 3 points then they won't be collinear.
Take any 10 points in a plane and name them as A, B ,C ,D ,E ,F ,G ,H ,I ,J.
Now next statement says 4 of these points are joined to 6 of the remaining points. So divide these 10 points into two groups containing 4 and 6 points arbitrarily. Let's assume first group as A, B, C, D and second group contains remaining 6 points. Now each point in first group is joined to each point in second group.
So line segments formed are 6+6+6+6=6*4=24. Because A is joined to 6 points, B is joined to 6 points and so on.
Next statement says each point from the group of 6 is joined to 5 other points (these 5 points can be in any group). Now these 6 points are already connected to 4 points in other group, so we need to connect each point to 1 more point from the same group. So divide these group of six into 3 groups of 2 and each group of 2 will form one line segment. For example E and F, G and H, I and J. Here if any point is repeated in group of two, then it will be joined to more than 5 points so we don't repeat. Now the given condition is satisfied.
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Answer:
There are 10 different points such that no three of them don't lie on straight line. This means if you select any 3 points then they won't be collinear.
Take any 10 points in a plane and name them as A, B ,C ,D ,E ,F ,G ,H ,I ,J.
Now next statement says 4 of these points are joined to 6 of the remaining points. So divide these 10 points into two groups containing 4 and 6 points arbitrarily. Let's assume first group as A, B, C, D and second group contains remaining 6 points. Now each point in first group is joined to each point in second group.
So line segments formed are 6+6+6+6=6*4=24. Because A is joined to 6 points, B is joined to 6 points and so on.
Next statement says each point from the group of 6 is joined to 5 other points (these 5 points can be in any group). Now these 6 points are already connected to 4 points in other group, so we need to connect each point to 1 more point from the same group. So divide these group of six into 3 groups of 2 and each group of 2 will form one line segment. For example E and F, G and H, I and J. Here if any point is repeated in group of two, then it will be joined to more than 5 points so we don't repeat. Now the given condition is satisfied.
So answer is 24+3=27.