# Number of possible triangles

Given six line segments of length 2, 3, 4, 5, 6, 7 units, the number of triangles that can be formed by these segments is

$(A)^6C_3 – 7$

$(B)^6C_3 – 6$

$(C)^6C_3– 5$

$(D)^6C_3 – 4$

I know that the no. of ways in which I can make triangles from these lines is 6C3 since their are six lines and I will have to use 3 lines at a time. But, I will have to minus the cases were the chosen segments will just result in a straight line and not a triangle. How can I do that?

• You also have to exclude cases like $(2,3,7)$ where you cannot even make a straight line. What the problem asks you is essentially just to count the number of such "impossible" combinations in whichever way you want. You don't need to be particularly smart about it. – Henning Makholm Sep 13 '13 at 13:10

We have the numbers $2,3,4,5,6,7$. Recall that the sum of any two sides of a triangle must be greater than the third side. Equivalently, the sum of the two shortest sides must be greater than the third side.

We are making a triangle. The shortest stick we use can be any of $2,3,4,5$. We find for each choice of shortest stick, how many bad choices there are for the other two sticks. (Bad choice = No triangle.)

Shortest stick is 2: If next shortest stick is $3$, then $5$, $6$, and $7$ for third are bad.

If next shortest after $2$ is $4$, then $5$ and $6$ for third are bad.

If next shortest after $2$ is $5$, then $7$ for third is bad.

Total bad here: $3+2+1=6$.

Shortest stick is 3: You can count the number of bads of this type.

Shortest is 4 or 5: No bads.

Now conclusion can be reached.

• There are seven cases where a triangle cannot be formed.Therefore, I think the answer is (A). – Rajath Krishna R Sep 13 '13 at 13:27