There are 4, 6, and 7 points on three lines. How many quadrilaterals is it possible to create with given points as vertices?

  1707

Step-by-step explanation:

Let's designate the three sets of collinear points, A, B, C, having 4, 6, 7 points, respectively.

Since there are 3 sets of collinear points, exactly two of the vertices must come from the same set.

For two vertices from set A, the remaining two must come from the 13 members of sets B and C. There are a total of (4C2)(13C2) = 468 such quadrilaterals.

For two vertices from set B, we have already counted the quadrilaterals that result when the remaining two are from set A. There are 4·7 = 28 ways to have one each from sets A and C, and 7C2 = 21 ways to have two from set C. Thus, the additional number of quadrilaterals having 2 vertices in set B is ...

  (6C2)(28 +21) = 735

For two vertices from set C, we have already counted the cases where two are from A or two are from B. There are 4·6 = 24 ways to have one each of the remaining vertices from sets A and B. Then the number of additional quadrilaterals having two points from set C is ...

  (7C2)(4)(6) = 504

So, the total number of unique quadrilaterals is ...

  468 +735 +504 = 1707

__

nCk means "the number of ways to choose k from n"

nCk = n!/(k!(n-k)!)

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

step-by-step explanation: