6 POINTS (A,B,C,D,E,F) LIE ON A CIRCLE

- anonymous

6 POINTS (A,B,C,D,E,F) LIE ON A CIRCLE

- chestercat

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- anonymous

That's cool. Also, nice caps.

- anonymous

|dw:1342406130806:dw|

- anonymous

Find the maximum no. of Points of Intersection of chords inside the circle?
LOL @SmoothMath !

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## More answers

- anonymous

fe || ad || bc ?

- anonymous

huh?

- anonymous

Virtus, why not draw in all of the possible chords?

- anonymous

|dw:1342406517682:dw|

- anonymous

oh then just count it

- anonymous

Truedat, sab.

- anonymous

no but you are supposed to use combinations to solve it, Like in the exam you can't just say refer to the diagram. You actually have to show them how you can get the answer through combinations!

- UnkleRhaukus

well for example (a) can join with 5 other point to make a cord,

- anonymous

Is there any statement about where the points are on the circle?

- anonymous

no there isn't
but the answer is 6C4 in case you were wondering

- UnkleRhaukus

i got \[^6C_2\]

- UnkleRhaukus

\[^6C_2=15=^6C_4\]

- anonymous

@UnkleRhaukus
but then there was a question prior to that that asked How many chords can be drawn by joining these different pairs of points and the answer was 6C2
Why couldn't they have written 6C2 for the second question?

- UnkleRhaukus

hmm im not sure

- anonymous

do you know where the 4 may have come from then in 6C4?

- anonymous

because i think i understand if it is 2, cause its kinda like that hand shake question then

- anonymous

ncr = nc(n-r)

- UnkleRhaukus

is suppose the question is asking the number of intersections the cords make

- anonymous

huh?

- anonymous

i guess we should calculate # intersections per one point, and multiply it with 6
then we are using counting... not combintions lol

- anonymous

I honestly don't see how it's a combination question, but hey.

- anonymous

im seeing a pattern maybe we can work backward
# intersecting points for point "a" inside the circle : 1.3 + 2.2 + 3.1 = 10
total # of intersecting points possible = 10*6 = 60

- anonymous

thats the answer. lets see if the pattern means something in combination terms
1.3 + 2.2 + 3.1
r(n) + (r+1)(n-1) + (r+2)(n-2) + ....

- anonymous

thats same as
SUM (r)(n-r)

- anonymous

|dw:1342408544273:dw|

- anonymous

this looks more compact than any combination formula... but it would be good to see if this can be converted to NCR s...

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