mathmath333
  • mathmath333
There are 25 points on a plane of which 7 are collinear . How many quadrilaterals can be formed from these points ?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{ There are 25 points on a plane of which 7 are collinear . }\hspace{.33em}\\~\\ & \normalsize \text{ How many quadrilaterals can be formed from these points ?}\hspace{.33em}\\~\\ \end{align}}\)
mathmate
  • mathmate
So there are two groups of points, 7 collinear, and 18 non-colinnear.
mathmath333
  • mathmath333
yes

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mathmate
  • mathmate
And there cannot be 3 or more points in a quad which are collinear.
mathmate
  • mathmate
See if you can think along those lines. I'll be back.
mathmath333
  • mathmath333
18C4
jim_thompson5910
  • 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.
dan815
  • dan815
|dw:1441331618617:dw|
dan815
  • dan815
i think jim has the right idea
dan815
  • dan815
18 choose 4 + 18 choose 2 * 7 choose 2 + 18 choose 3 * 7 choose 1
jim_thompson5910
  • jim_thompson5910
I agree with `18 choose 4 + 18 choose 2 * 7 choose 2 + 18 choose 3 * 7 choose 1` use this formula \[\Large _n C _r = \frac{n!}{r!(n-r)!}\]
mathmate
  • 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
mathmath333
  • mathmath333
u didnt add ----18 choose 1 * 7 choose 3 ---- ??
dan815
  • dan815
for k colinnear and n total points to be |dw:1441331943171:dw|
dan815
  • dan815
who doesnt like general formulas hehe
beginnersmind
  • beginnersmind
For some collection of 4 points you can form more than one quadrilatereal.|dw:1441332129959:dw|
mathmate
  • mathmate
Oh, sorry, forget the denominator. That's for probability!
jim_thompson5910
  • 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"
dan815
  • dan815
oh that is true :O
mathmath333
  • mathmath333
oh i see that thanks for that
mathmate
  • mathmate
\(\Large \left(\begin{matrix}A \\ i\end{matrix}\right)\left(\begin{matrix}B \\ n-i\end{matrix}\right)\)
jim_thompson5910
  • jim_thompson5910
based on what beginnersmind drew, it looks like order matters
dan815
  • dan815
i think they didnt consider than when they asked the question tbh lol
dan815
  • dan815
cause it really confuses the issue, on how many quads are possible with the given set of 4 arb points
beginnersmind
  • beginnersmind
@jim_thompson5910 It depends on the exact position of the points. For most it doesn't.
mathmate
  • 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.
mathmath333
  • mathmath333
i got 11985
dan815
  • dan815
are there answer choices?
mathmath333
  • mathmath333
yes 11985 is an answer choice
dan815
  • dan815
give me the other ones too just in case
dan815
  • dan815
is there anything that is exactly 3 times that?
dan815
  • dan815
or some multiple of that
mathmath333
  • mathmath333
a.) 5206 b.) 2603 c.) 13015 d.) 11985
mathmate
  • mathmate
Yes, I have 11985.
dan815
  • dan815
okay u r done
mathmath333
  • mathmath333
thnks all

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