How many ordered triples of integers $(a, b, c)$ satisfy the following system?
$$
\begin{cases} ab + c &= 17 \\ a + bc &= 19 \end{cases}
$$
$$
\mathrm a. ~ 2\qquad \mathrm b.~3\qquad \mathrm c. ~4 \qquad \mathrm d. ~5 \qquad \mathrm e. ~6
$$
To solve the system of equations:
\[
\begin{cases}
ab + c &= 17 \\
a + bc &= 19
\end{cases}
\]

1. **Subtract the two equations:**
   \[
   (ab + c) - (a + bc) = 17 - 19
   \]
   Simplifying, we get:
   \[
   ab + c - a - bc = -2
   \]
   Rearrange terms:
   \[
   a(b-1) - c(b-1) = -2
   \]
   Factor out \((b-1)\):
   \[
   (a - c)(b - 1) = -2
   \]

2. **Add the two equations:**
   \[
   (ab + c) + (a + bc) = 17 + 19
   \]
   Simplifying, we get:
   \[
   ab + c + a + bc = 36
   \]
   Rearrange terms:
   \[
   a(1 + b) + c(1 + b) = 36
   \]
   Factor out \((1 + b)\):
   \[
   (a + c)(b + 1) = 36
   \]

3. **Analyze the factorizations:**
   We now have two key equations:
   \[
   (a - c)(b - 1) = -2
   \]
   \[
   (a + c)(b + 1) = 36
   \]

4. **Find integer solutions:**
   We need to find integer values of \(a\), \(b\), and \(c\) that satisfy both equations. We will consider the possible factorizations of \(-2\) and \(36\).

   - For \((a - c)(b - 1) = -2\), the possible pairs \((a - c, b - 1)\) are:
     \[
     (-1, 2), (1, -2), (-2, 1), (2, -1)
     \]

   - For \((a + c)(b + 1) = 36\), the possible pairs \((a + c, b + 1)\) are:
     \[
     (1, 36), (2, 18), (3, 12), (4, 9), (6, 6), (-1, -36), (-2, -18), (-3, -12), (-4, -9), (-6, -6)
     \]

5. **Match pairs:**
   We need to find pairs \((a, c)\) and \(b\) that satisfy both equations. We will test each pair:

   - For \((a - c, b - 1) = (-1, 2)\):
     \[
     b = 3 \quad \text{and} \quad (a + c)(3 + 1) = 36 \implies (a + c)4 = 36 \implies a + c = 9
     \]
     \[
     a - c = -1
     \]
     Solving these:
     \[
     a + c = 9 \quad \text{and} \quad a - c = -1
     \]
     Adding these:
     \[
     2a = 8 \implies a = 4
     \]
     Subtracting these:
     \[
     2c = 10 \implies c = 5
     \]
     So, \((a, b, c) = (4, 3, 5)\).

   - For \((a - c, b - 1) = (1, -2)\):
     \[
     b = -1 \quad \text{and} \quad (a + c)(-1 + 1) = 36 \implies (a + c)0 = 36 \implies \text{no solution}
     \]

   - For \((a - c, b - 1) = (-2, 1)\):
     \[
     b = 2 \quad \text{and} \quad (a + c)(2 + 1) = 36 \implies (a + c)3 = 36 \implies a + c = 12
     \]
     \[
     a - c = -2
     \]
     Solving these:
     \[
     a + c = 12 \quad \text{and} \quad a - c = -2
     \]
     Adding these:
     \[
     2a = 10 \implies a = 5
     \]
     Subtracting these:
     \[
     2c = 14 \implies c = 7
     \]
     So, \((a, b, c) = (5, 2, 7)\).

   - For \((a - c, b - 1) = (2, -1)\):
     \[
     b = 0 \quad \text{and} \quad (a + c)(0 + 1) = 36 \implies a + c = 36
     \]
     \[
     a - c = 2
     \]
     Solving these:
     \[
     a + c = 36 \quad \text{and} \quad a - c = 2
     \]
     Adding these:
     \[
     2a = 38 \implies a = 19
     \]
     Subtracting these:
     \[
     2c = 34 \implies c = 17
     \]
     So, \((a, b, c) = (19, 0, 17)\).

6. **Conclusion:**
   The valid ordered triples \((a, b, c)\) are:
   \[
   (4, 3, 5), (5, 2, 7), (19, 0, 17)
   \]

The final answer is \(\boxed{3}\)