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SmoothMathBest ResponseYou've already chosen the best response.3
That's cool. Also, nice caps.
 one year ago

virtusBest ResponseYou've already chosen the best response.0
Find the maximum no. of Points of Intersection of chords inside the circle? LOL @SmoothMath !
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.3
Virtus, why not draw in all of the possible chords?
 one year ago

sabeelioBest ResponseYou've already chosen the best response.0
oh then just count it
 one year ago

virtusBest ResponseYou've already chosen the best response.0
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

UnkleRhaukusBest ResponseYou've already chosen the best response.0
well for example (a) can join with 5 other point to make a cord,
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.3
Is there any statement about where the points are on the circle?
 one year ago

virtusBest ResponseYou've already chosen the best response.0
no there isn't but the answer is 6C4 in case you were wondering
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[^6C_2=15=^6C_4\]
 one year ago

virtusBest 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?
 one year ago

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

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

UnkleRhaukusBest ResponseYou've already chosen the best response.0
is suppose the question is asking the number of intersections the cords make
 one year ago

ramya25Best ResponseYou've already chosen the best response.1
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

SmoothMathBest ResponseYou've already chosen the best response.3
I honestly don't see how it's a combination question, but hey.
 one year ago

ramya25Best ResponseYou've already chosen the best response.1
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

ramya25Best ResponseYou've already chosen the best response.1
thats 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) + ....
 one year ago

ramya25Best ResponseYou've already chosen the best response.1
thats same as SUM (r)(nr)
 one year ago

ramya25Best ResponseYou've already chosen the best response.1
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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