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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 ?

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{ There are 25 points on a plane of which 7 are collinear . }\hspace{.33em}\\~\\ & \normalsize \text{ How many quadrilaterals can be formed from these points ?}\hspace{.33em}\\~\\ \end{align}}\)

  2. mathmate
    • one year ago
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    So there are two groups of points, 7 collinear, and 18 non-colinnear.

  3. mathmath333
    • one year ago
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    yes

  4. mathmate
    • one year ago
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    And there cannot be 3 or more points in a quad which are collinear.

  5. mathmate
    • one year ago
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    See if you can think along those lines. I'll be back.

  6. mathmath333
    • one year ago
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    18C4

  7. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    |dw:1441331618617:dw|

  9. dan815
    • one year ago
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    i think jim has the right idea

  10. dan815
    • one year ago
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    18 choose 4 + 18 choose 2 * 7 choose 2 + 18 choose 3 * 7 choose 1

  11. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    u didnt add ----18 choose 1 * 7 choose 3 ---- ??

  14. dan815
    • one year ago
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    for k colinnear and n total points to be |dw:1441331943171:dw|

  15. dan815
    • one year ago
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    who doesnt like general formulas hehe

  16. beginnersmind
    • one year ago
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    For some collection of 4 points you can form more than one quadrilatereal.|dw:1441332129959:dw|

  17. mathmate
    • one year ago
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    Oh, sorry, forget the denominator. That's for probability!

  18. jim_thompson5910
    • one year ago
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    `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
    • one year ago
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    oh that is true :O

  20. mathmath333
    • one year ago
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    oh i see that thanks for that

  21. mathmate
    • one year ago
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    \(\Large \left(\begin{matrix}A \\ i\end{matrix}\right)\left(\begin{matrix}B \\ n-i\end{matrix}\right)\)

  22. jim_thompson5910
    • one year ago
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    based on what beginnersmind drew, it looks like order matters

  23. dan815
    • one year ago
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    i think they didnt consider than when they asked the question tbh lol

  24. dan815
    • one year ago
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    cause it really confuses the issue, on how many quads are possible with the given set of 4 arb points

  25. beginnersmind
    • one year ago
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    @jim_thompson5910 It depends on the exact position of the points. For most it doesn't.

  26. mathmate
    • one year ago
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    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
    • one year ago
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    i got 11985

  28. dan815
    • one year ago
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    are there answer choices?

  29. mathmath333
    • one year ago
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    yes 11985 is an answer choice

  30. dan815
    • one year ago
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    give me the other ones too just in case

  31. dan815
    • one year ago
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    is there anything that is exactly 3 times that?

  32. dan815
    • one year ago
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    or some multiple of that

  33. mathmath333
    • one year ago
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    a.) 5206 b.) 2603 c.) 13015 d.) 11985

  34. mathmate
    • one year ago
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    Yes, I have 11985.

  35. dan815
    • one year ago
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    okay u r done

  36. mathmath333
    • one year ago
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    thnks all

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