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

- mathmath333

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- 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

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

- mathmath333

yes

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## More answers

- mathmate

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

- mathmate

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

- mathmath333

18C4

- 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

|dw:1441331618617:dw|

- dan815

i think jim has the right idea

- dan815

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

- 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

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

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

- dan815

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

- dan815

who doesnt like general formulas hehe

- beginnersmind

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

- mathmate

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

- 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

oh that is true :O

- mathmath333

oh i see that thanks for that

- mathmate

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

- jim_thompson5910

based on what beginnersmind drew, it looks like order matters

- dan815

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

- dan815

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

- beginnersmind

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

- 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

i got 11985

- dan815

are there answer choices?

- mathmath333

yes 11985 is an answer choice

- dan815

give me the other ones too just in case

- dan815

is there anything that is exactly 3 times that?

- dan815

or some multiple of that

- mathmath333

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

- mathmate

Yes, I have 11985.

- dan815

okay u r done

- mathmath333

thnks all

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