Another one. A dice is rolled thrice. The numbers appearing are noted and if possible, Constructed a triangle with. Find the ratio of the number of Scalene tringles possible to isosceles triangles that could be made.
I have calculated quite roughly. But is it something like 22/107? This could be very stupid, but the hazards are all mine.
Ummmmm. No. Numerator is either 5 or 7. Not sure. Not 22.
Given answer for Numerator is 5. (Not sure about accuracy) Still, basic condition you applied?
there are 6 choose 3 or 20 combinations for possible scalene triangles only 6 will work 2,3,4 2,4,5 2,5,6 3,4,5 3,4,6 4,5,6 for isosceles triangles, there are 30 possibilities 21 will work, 9 will not -these will not anything with two 1's 2,2,4 2,2,5 2,2,6 3,3,6 --> ratio = 6/21 = 2/7
So the denominator may be correct? I posted for ratio isosceles to scalene. I have another answer that is 7 to 22 (scalene to isosceles)
@dumbcow 5,3,6 too, right? And Did you employ trial and Error or something else?
Apporv. 7 seems correct. I'm getting 21 for isosceles.
See the isosceles possibilities according to me are: 221 223 332 332 334 335 441 442 443 443 445 446 and same way 5 possibilities each for the 55.. and 66.. series) so that is. '21" (i had miscalculated it as 22 earlier) Now for scalene triplets, i list the ones which are not possible: 123 124 125 126 134 135 136 145 146 156 235 236 that is about 12 now no. of ways of choosing 3 nos. out of 6 is 6C3. that is 20. minus the no. of possibilities. so 20-12 = 8 so, scalene to isosceles =8/21
Know that's a very crude method, and I may have missed out a couple, according to dumbcow's answer.
Also 6C3 Doesnt apply here. You have a total 6*6*6 = 216 possibilities.
oh yeah another that is not possible in the scalene part "246" so it becomes (20-13)/21 =7/21
Ohhhhhh wait.
You reduced it without that in the first step. Yes that works 6 C3
6C3 yes. am applying them for "scalene triplets". that is abc where a b and c none are the same.
Thanksguys. So 7/21 it is. Though it took time right? And You just have too use a+b>c by trial and error? Thats just too long. Something shorter to cut the time short?
Hmm, yes if you are a master of PnC. (Though it didn't take much time as I counted pretty systematically, but accuracy is always an issue then.)
Exactly. I tred Integral solutions usng a dummy varible. Didnt help too much.
oh sorry guys, yeah i missed one ---> 7/21 it is and yes i just counted which combinations would work one-by-one 30 possible isosceles 20 possible scalene
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