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virtus

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

  • one year ago
  • one year ago

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

    • one year ago
  2. virtus
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    |dw:1342406130806:dw|

    • one year ago
  3. virtus
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    Find the maximum no. of Points of Intersection of chords inside the circle? LOL @SmoothMath !

    • one year ago
  4. ramya25
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    fe || ad || bc ?

    • one year ago
  5. virtus
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    huh?

    • one year ago
  6. SmoothMath
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    Virtus, why not draw in all of the possible chords?

    • one year ago
  7. virtus
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    |dw:1342406517682:dw|

    • one year ago
  8. sabeelio
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    oh then just count it

    • one year ago
  9. SmoothMath
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    Truedat, sab.

    • one year ago
  10. virtus
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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!

    • one year ago
  11. UnkleRhaukus
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    well for example (a) can join with 5 other point to make a cord,

    • one year ago
  12. SmoothMath
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    Is there any statement about where the points are on the circle?

    • one year ago
  13. virtus
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    no there isn't but the answer is 6C4 in case you were wondering

    • one year ago
  14. UnkleRhaukus
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    i got \[^6C_2\]

    • one year ago
  15. UnkleRhaukus
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    \[^6C_2=15=^6C_4\]

    • one year ago
  16. virtus
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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?

    • one year ago
  17. UnkleRhaukus
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    hmm im not sure

    • one year ago
  18. virtus
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    do you know where the 4 may have come from then in 6C4?

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

    • one year ago
  20. ramya25
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    ncr = nc(n-r)

    • one year ago
  21. UnkleRhaukus
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    is suppose the question is asking the number of intersections the cords make

    • one year ago
  22. virtus
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    huh?

    • one year ago
  23. ramya25
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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

    • one year ago
  24. SmoothMath
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    I honestly don't see how it's a combination question, but hey.

    • one year ago
  25. ramya25
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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

    • one year ago
  26. ramya25
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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) + ....

    • one year ago
  27. ramya25
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    thats same as SUM (r)(n-r)

    • one year ago
  28. ramya25
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    |dw:1342408544273:dw|

    • one year ago
  29. ramya25
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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...

    • one year ago
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