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virtus

  • 2 years ago

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

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  1. SmoothMath
    • 2 years ago
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    That's cool. Also, nice caps.

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

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

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

  5. virtus
    • 2 years ago
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    huh?

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

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

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

  9. SmoothMath
    • 2 years ago
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    Truedat, sab.

  10. virtus
    • 2 years ago
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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!

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

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

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

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

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

  16. virtus
    • 2 years ago
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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?

  17. UnkleRhaukus
    • 2 years ago
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    hmm im not sure

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

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

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

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

  22. virtus
    • 2 years ago
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    huh?

  23. ramya25
    • 2 years ago
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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

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

  25. ramya25
    • 2 years ago
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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

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

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

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

  29. ramya25
    • 2 years ago
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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...

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