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anonymous
 4 years ago
6 POINTS (A,B,C,D,E,F) LIE ON A CIRCLE
anonymous
 4 years ago
6 POINTS (A,B,C,D,E,F) LIE ON A CIRCLE

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's cool. Also, nice caps.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342406130806:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Find the maximum no. of Points of Intersection of chords inside the circle? LOL @SmoothMath !

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Virtus, why not draw in all of the possible chords?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342406517682:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh then just count it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no 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
 4 years ago
Best ResponseYou've already chosen the best response.0well for example (a) can join with 5 other point to make a cord,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is there any statement about where the points are on the circle?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no there isn't but the answer is 6C4 in case you were wondering

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[^6C_2=15=^6C_4\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you know where the 4 may have come from then in 6C4?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because i think i understand if it is 2, cause its kinda like that hand shake question then

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0is suppose the question is asking the number of intersections the cords make

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i guess we should calculate # intersections per one point, and multiply it with 6 then we are using counting... not combintions lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I honestly don't see how it's a combination question, but hey.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im 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
 4 years ago
Best ResponseYou've already chosen the best response.0thats the answer. lets see if the pattern means something in combination terms 1.3 + 2.2 + 3.1 r(n) + (r+1)(n1) + (r+2)(n2) + ....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats same as SUM (r)(nr)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342408544273:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this 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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