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The total number of non congruent integer sided triangles whose sides belong to the set{$10,11,12,\cdots,22$} are---

For a triangle to be formed,sum of any two sided must be greater than the third side,however if 10,11 are the sides of the triangle then 10+11=21,so we cannot have 21 and 22 for the third side.Similarly if we had 10,12 as two of the sides,22 cannot be the third side.Other than this,there seems no restrictions on choosing the three sides.There are 9 options for third side if two sides are (10,11).similar is case for (10,12).Now for number of possible cases for 11 as one of the side is $\binom{11}{2}$(as we have already counted for number of cases with (11,10).Similar is case for 12 as one of the side.The other triangles can be formed by choosing any three sides from {$13,14,\cdots,22$}. hence my answer for total number of scalene triangles formed is-- $$9+9+\binom{11}{2}+\binom{11}{2}+\binom{10}{3}$$It is easy to calculate the number of isosceles triangles and equilateral ones.However I am getting wrong answer.Can anyone suggest where I am wrong.Thanks.

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  • $\begingroup$ In your second paragraph it would be better to say you are counting scalene triangles at the start. $\endgroup$ – Ross Millikan Nov 27 '16 at 15:45
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For scalene triangles, the easiest way is to note that you can choose any three sides of the thirteen available except $(10,11,21), (10,11,22), (10,12,22)$ so the count is ${13 \choose 3}-3=283$. You have counted $(11,12,20)$, for example, twice, once as one of the $11$s and once as one of the $12$s. You have missed counting the triangles with $10$ as one side unless the second side is $11$ or $12$. If yo change your second $11 \choose 2$ to be $10 \choose 2$ to skip the double count of $11,12$ and add $10 choose 2$ for $10$ plus two sides greater than $12$, you again get $283$

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