Find the Number of non-congruent triangles (integer sided) whose sides belong to the set


This is a question in a national level based examination. Now in here what i can see that be taking each of the numbers as largest side(except for 10 and 11) we can find the number of triangles thus formed in each case by a brute force method.

Now also the question says non congruent triangles which makes the problem even more complex.Now do we apply recurssion here or simply this a problem of permutations and combinations.

Also i accept that my method of a brute force approach is rather an unintelligent way to solve it,so i am looking for answer that can solve it in 5 lines or so.


1 Answer 1


You just need to choose three numbers from the set with replacement so that the sum of the smaller two is greater than the largest. The lengths of the sides determine the triangle. As there are $13$ numbers in the set, there are $13$ choices where the numbers are all the same, $13\cdot 12=156$ where two numbers are the same, and $\frac 16\cdot 13 \cdot 12 \cdot 11=286$ where all three numbers are different. Because $22$ is barely more than twice $10$ we can hand count the ones that do not form a triangle. There are none with all the sides the same. With two sides equal, the only failures are $(10,10,20), (10,10,21), (10,10,22), (11,11,22)$. With all sides different there is just $(10,11,21), (10,11,22),(10,12,22)$. There are therefore $13+156+286-7=448$ triangles that can be formed.

  • $\begingroup$ can you tell me the case of three different numbers $\endgroup$
    – Pole_Star
    Oct 25, 2017 at 7:41
  • $\begingroup$ it is 13 C3 right $\endgroup$
    – Pole_Star
    Oct 25, 2017 at 7:44
  • $\begingroup$ That is correct. For three different numbers you don't care what order you pick them in, so it is $13 \choose 3$. For two numbers you do care because you can decide to take two of the first and one of the second. $(13,13,12)$ is different from $(12,12,13)$ $\endgroup$ Oct 25, 2017 at 14:29

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