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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 ?
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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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{ There are 25 points on a plane of which 7 are collinear . }\hspace{.33em}\\~\\ & \normalsize \text{ How many quadrilaterals can be formed from these points ?}\hspace{.33em}\\~\\ \end{align}}\)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1So there are two groups of points, 7 collinear, and 18 noncolinnear.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1And there cannot be 3 or more points in a quad which are collinear.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1See if you can think along those lines. I'll be back.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2Let A,B,C,D,E,F,G be the 7 collinear points. Let H through Y be the other noncollinear 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
 one year ago
Best ResponseYou've already chosen the best response.1i think jim has the right idea

dan815
 one year ago
Best ResponseYou've already chosen the best response.118 choose 4 + 18 choose 2 * 7 choose 2 + 18 choose 3 * 7 choose 1

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2I agree with `18 choose 4 + 18 choose 2 * 7 choose 2 + 18 choose 3 * 7 choose 1` use this formula \[\Large _n C _r = \frac{n!}{r!(nr)!}\]

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1And use \(\Large \frac{\left(\begin{matrix}A \\ i\end{matrix}\right)\left(\begin{matrix}B \\ ni\end{matrix}\right)}{\left(\begin{matrix}A+B \\ n\end{matrix}\right)}\) A=18, B=7, n=4, i=2,3,4

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0u didnt add 18 choose 1 * 7 choose 3  ??

dan815
 one year ago
Best ResponseYou've already chosen the best response.1for k colinnear and n total points to be dw:1441331943171:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.1who doesnt like general formulas hehe

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1For some collection of 4 points you can form more than one quadrilatereal.dw:1441332129959:dw

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Oh, sorry, forget the denominator. That's for probability!

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2`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"

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0oh i see that thanks for that

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large \left(\begin{matrix}A \\ i\end{matrix}\right)\left(\begin{matrix}B \\ ni\end{matrix}\right)\)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2based on what beginnersmind drew, it looks like order matters

dan815
 one year ago
Best ResponseYou've already chosen the best response.1i think they didnt consider than when they asked the question tbh lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.1cause it really confuses the issue, on how many quads are possible with the given set of 4 arb points

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.1@jim_thompson5910 It depends on the exact position of the points. For most it doesn't.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1But we're just choosing the four points, order does not count, even though there is only one way to make the correct quad.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1are there answer choices?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0yes 11985 is an answer choice

dan815
 one year ago
Best ResponseYou've already chosen the best response.1give me the other ones too just in case

dan815
 one year ago
Best ResponseYou've already chosen the best response.1is there anything that is exactly 3 times that?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1or some multiple of that

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0a.) 5206 b.) 2603 c.) 13015 d.) 11985
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