For how many ordered triples $ (a,b,c)$ of positive integers are the equations $ abc\plus{}9 \equal{} ab\plus{}bc\plus{}ca$ and $ a\plus{}b\plus{}c \equal{} 10$ satisfied?
1. **Assume without loss of generality that \( a \geq b \geq c \).**

2. **Rewrite the given equation \( abc + 9 = ab + bc + ca \):**
   \[
   abc + 9 = ab + bc + ca
   \]
   Rearrange the terms:
   \[
   abc - ab - bc - ca = -9
   \]
   Factor by grouping:
   \[
   ab(c-1) + c(a+b) = -9
   \]

3. **Use the second equation \( a + b + c = 10 \):**
   \[
   a + b + c = 10
   \]

4. **Analyze the possible values for \( c \):**
   Since \( a \geq b \geq c \), we have \( c \leq \frac{10}{3} \approx 3.33 \). Therefore, \( c \) can be 1, 2, or 3.

5. **Case 1: \( c = 1 \):**
   Substitute \( c = 1 \) into the equations:
   \[
   a + b + 1 = 10 \implies a + b = 9
   \]
   \[
   ab(1-1) + 1(a+b) = -9 \implies a + b = 9
   \]
   This is consistent. The pairs \((a, b)\) that satisfy \( a + b = 9 \) are:
   \[
   (8, 1), (7, 2), (6, 3), (5, 4)
   \]
   Thus, the ordered triples are \((8, 1, 1), (7, 2, 1), (6, 3, 1), (5, 4, 1)\).

6. **Case 2: \( c = 2 \):**
   Substitute \( c = 2 \) into the equations:
   \[
   a + b + 2 = 10 \implies a + b = 8
   \]
   \[
   ab(2-1) + 2(a+b) = -9 \implies ab + 2(a + b) = -9
   \]
   Substitute \( a + b = 8 \):
   \[
   ab + 2 \cdot 8 = -9 \implies ab + 16 = -9 \implies ab = -25
   \]
   This is not possible since \( ab \) must be positive.

7. **Case 3: \( c = 3 \):**
   Substitute \( c = 3 \) into the equations:
   \[
   a + b + 3 = 10 \implies a + b = 7
   \]
   \[
   ab(3-1) + 3(a+b) = -9 \implies 2ab + 3 \cdot 7 = -9 \implies 2ab + 21 = -9 \implies 2ab = -30 \implies ab = -15
   \]
   This is not possible since \( ab \) must be positive.

8. **Conclusion:**
   The only valid solutions are from Case 1, where \( c = 1 \).

The final answer is \(\boxed{4}\).