2017 AMC 10A Problem 10

Note that no one side can be greater than or equal to the sum of the other side lengths. Let $x$ be the length fourth rod. Then we have that $$x \lt 3 + 7 + 15$$ and $$x + 3 + 7 \gt 15$$ Simplifying, we know that $$5\lt x\lt 25.$$ Counting the number of integers in this range, we are left with $25 - 5 - 1 = 19$ values for $x.$ The rods with length $7$ and $15$ are already being used, however, so $x$ cannot equal these. This leaves $19 - 2 = 17$ viable solutions for $x.$ Thus, B is the correct answer.