mestrec wrote:
Bunuel, is there a easier way to do the math, instead of multiplying 12 x 11 x 10 x 9 x 8 x 7.....?
As far I understood your solution, I would need to do that at the C 3 12 combination.
Regards and have a great year!
Bunuel wrote:
Bumping for review and further discussion.
Hi,
Iwillhelp you out on that ..
the Q basically says there are three lines with 3,4,5 points respectively..
lets find ways triangle can be formed..
triangle requires three points,..
1) 2 points from line containing 3 points and one from remaining 4+5 points=3C2*(4+5)=27..
2) 2 points from line containing 4 points and one from remaining 3+5 points=4C2*(3+5)=48..
3) 2 points from line containing 5 points and one from remaining 4+3 points=5C2*(4+3)=70..
4) 1 point from each line containing 3,4,5 points =3*4*5=60..
total 27+48+70+60=205..
also you do not have to do the entire calculation in 12C3..
12C3=12!/3!9!= 12*11*10*9!/3!9!=12*11*10/3*2...
so you see you just have to multiply 12,11,and 10..
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