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That's cool. Also, nice caps.
Find the maximum no. of Points of Intersection of chords inside the circle? LOL @SmoothMath !
fe || ad || bc ?
Virtus, why not draw in all of the possible chords?
oh then just count it
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!
well for example (a) can join with 5 other point to make a cord,
Is there any statement about where the points are on the circle?
no there isn't but the answer is 6C4 in case you were wondering
i got \[^6C_2\]
@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?
hmm im not sure
do you know where the 4 may have come from then in 6C4?
because i think i understand if it is 2, cause its kinda like that hand shake question then
ncr = nc(n-r)
is suppose the question is asking the number of intersections the cords make
i guess we should calculate # intersections per one point, and multiply it with 6 then we are using counting... not combintions lol
I honestly don't see how it's a combination question, but hey.
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
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) + ....
thats same as SUM (r)(n-r)
this looks more compact than any combination formula... but it would be good to see if this can be converted to NCR s...