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?  

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.
