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virtus Group Title

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

  • 2 years ago
  • 2 years ago

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  1. SmoothMath Group Title
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    That's cool. Also, nice caps.

    • 2 years ago
  2. virtus Group Title
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    |dw:1342406130806:dw|

    • 2 years ago
  3. virtus Group Title
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    Find the maximum no. of Points of Intersection of chords inside the circle? LOL @SmoothMath !

    • 2 years ago
  4. ramya25 Group Title
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    fe || ad || bc ?

    • 2 years ago
  5. virtus Group Title
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    huh?

    • 2 years ago
  6. SmoothMath Group Title
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    Virtus, why not draw in all of the possible chords?

    • 2 years ago
  7. virtus Group Title
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    |dw:1342406517682:dw|

    • 2 years ago
  8. sabeelio Group Title
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    oh then just count it

    • 2 years ago
  9. SmoothMath Group Title
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    Truedat, sab.

    • 2 years ago
  10. virtus Group Title
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    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!

    • 2 years ago
  11. UnkleRhaukus Group Title
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    well for example (a) can join with 5 other point to make a cord,

    • 2 years ago
  12. SmoothMath Group Title
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    Is there any statement about where the points are on the circle?

    • 2 years ago
  13. virtus Group Title
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    no there isn't but the answer is 6C4 in case you were wondering

    • 2 years ago
  14. UnkleRhaukus Group Title
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    i got \[^6C_2\]

    • 2 years ago
  15. UnkleRhaukus Group Title
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    \[^6C_2=15=^6C_4\]

    • 2 years ago
  16. virtus Group Title
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    @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?

    • 2 years ago
  17. UnkleRhaukus Group Title
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    hmm im not sure

    • 2 years ago
  18. virtus Group Title
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    do you know where the 4 may have come from then in 6C4?

    • 2 years ago
  19. virtus Group Title
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    because i think i understand if it is 2, cause its kinda like that hand shake question then

    • 2 years ago
  20. ramya25 Group Title
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    ncr = nc(n-r)

    • 2 years ago
  21. UnkleRhaukus Group Title
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    is suppose the question is asking the number of intersections the cords make

    • 2 years ago
  22. virtus Group Title
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    huh?

    • 2 years ago
  23. ramya25 Group Title
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    i guess we should calculate # intersections per one point, and multiply it with 6 then we are using counting... not combintions lol

    • 2 years ago
  24. SmoothMath Group Title
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    I honestly don't see how it's a combination question, but hey.

    • 2 years ago
  25. ramya25 Group Title
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    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

    • 2 years ago
  26. ramya25 Group Title
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    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) + ....

    • 2 years ago
  27. ramya25 Group Title
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    thats same as SUM (r)(n-r)

    • 2 years ago
  28. ramya25 Group Title
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    |dw:1342408544273:dw|

    • 2 years ago
  29. ramya25 Group Title
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    this looks more compact than any combination formula... but it would be good to see if this can be converted to NCR s...

    • 2 years ago
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