## mathmath333 one year ago There are 25 points on a plane of which 7 are collinear . How many quadrilaterals can be formed from these points ?

1. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{ There are 25 points on a plane of which 7 are collinear . }\hspace{.33em}\\~\\ & \normalsize \text{ How many quadrilaterals can be formed from these points ?}\hspace{.33em}\\~\\ \end{align}}

2. mathmate

So there are two groups of points, 7 collinear, and 18 non-colinnear.

3. mathmath333

yes

4. mathmate

And there cannot be 3 or more points in a quad which are collinear.

5. mathmate

See if you can think along those lines. I'll be back.

6. mathmath333

18C4

7. jim_thompson5910

Let A,B,C,D,E,F,G be the 7 collinear points. Let H through Y be the other non-collinear points. Maybe break it up into cases Case 1: All four points are chosen from the set {H, I, J, ..., X, Y} Case 2: Three points are chosen from the set {H, I, J, ..., X, Y} while one point is chosen from {A,B,C,D,E,F,G} Case 3: Two points are chosen from the set {H, I, J, ..., X, Y} and two points are chosen from {A,B,C,D,E,F,G} It's impossible to have 1 point chosen from the set {H, I, J, ..., X, Y} and have 3 points chosen from {A,B,C,D,E,F,G} because you'd have a triangle forming instead of a quadrilateral. You cannot choose 4 points from {A,B,C,D,E,F,G} because you'd have a line only.

8. dan815

|dw:1441331618617:dw|

9. dan815

i think jim has the right idea

10. dan815

18 choose 4 + 18 choose 2 * 7 choose 2 + 18 choose 3 * 7 choose 1

11. jim_thompson5910

I agree with 18 choose 4 + 18 choose 2 * 7 choose 2 + 18 choose 3 * 7 choose 1 use this formula $\Large _n C _r = \frac{n!}{r!(n-r)!}$

12. mathmate

And use $$\Large \frac{\left(\begin{matrix}A \\ i\end{matrix}\right)\left(\begin{matrix}B \\ n-i\end{matrix}\right)}{\left(\begin{matrix}A+B \\ n\end{matrix}\right)}$$ A=18, B=7, n=4, i=2,3,4

13. mathmath333

u didnt add ----18 choose 1 * 7 choose 3 ---- ??

14. dan815

for k colinnear and n total points to be |dw:1441331943171:dw|

15. dan815

who doesnt like general formulas hehe

16. beginnersmind

For some collection of 4 points you can form more than one quadrilatereal.|dw:1441332129959:dw|

17. mathmate

Oh, sorry, forget the denominator. That's for probability!

18. jim_thompson5910

u didnt add ----18 choose 1 * 7 choose 3 ---- ?? @mathmath333 like I said, "It's impossible to have 1 point chosen from the set {H, I, J, ..., X, Y} and have 3 points chosen from {A,B,C,D,E,F,G} because you'd have a triangle forming instead of a quadrilateral"

19. dan815

oh that is true :O

20. mathmath333

oh i see that thanks for that

21. mathmate

$$\Large \left(\begin{matrix}A \\ i\end{matrix}\right)\left(\begin{matrix}B \\ n-i\end{matrix}\right)$$

22. jim_thompson5910

based on what beginnersmind drew, it looks like order matters

23. dan815

i think they didnt consider than when they asked the question tbh lol

24. dan815

cause it really confuses the issue, on how many quads are possible with the given set of 4 arb points

25. beginnersmind

@jim_thompson5910 It depends on the exact position of the points. For most it doesn't.

26. mathmate

But we're just choosing the four points, order does not count, even though there is only one way to make the correct quad.

27. mathmath333

i got 11985

28. dan815

29. mathmath333

yes 11985 is an answer choice

30. dan815

give me the other ones too just in case

31. dan815

is there anything that is exactly 3 times that?

32. dan815

or some multiple of that

33. mathmath333

a.) 5206 b.) 2603 c.) 13015 d.) 11985

34. mathmate

Yes, I have 11985.

35. dan815

okay u r done

36. mathmath333

thnks all