Find the number of ordered pairs of integers $(a, b)$ that satisfy the inequality

\[
1 < a < b+2 < 10.
\]

[i]Proposed by Lewis Chen
[/i]
1. We start with the given inequality:
   \[
   1 < a < b+2 < 10.
   \]
   We can break this into two parts:
   \[
   1 < a \quad \text{and} \quad a < b+2 < 10.
   \]

2. From \(a < b+2 < 10\), we can derive:
   \[
   a < b+2 \quad \text{and} \quad b+2 < 10.
   \]
   Simplifying \(b+2 < 10\), we get:
   \[
   b < 8.
   \]
   Therefore, \(b\) can take integer values from 1 to 7.

3. For each value of \(b\), we need to find the range of \(a\) such that \(1 < a < b+2\). Let's consider each possible value of \(b\):

   - If \(b = 1\):
     \[
     1 < a < 3 \implies a = 2 \quad (\text{1 value})
     \]
   - If \(b = 2\):
     \[
     1 < a < 4 \implies a = 2, 3 \quad (\text{2 values})
     \]
   - If \(b = 3\):
     \[
     1 < a < 5 \implies a = 2, 3, 4 \quad (\text{3 values})
     \]
   - If \(b = 4\):
     \[
     1 < a < 6 \implies a = 2, 3, 4, 5 \quad (\text{4 values})
     \]
   - If \(b = 5\):
     \[
     1 < a < 7 \implies a = 2, 3, 4, 5, 6 \quad (\text{5 values})
     \]
   - If \(b = 6\):
     \[
     1 < a < 8 \implies a = 2, 3, 4, 5, 6, 7 \quad (\text{6 values})
     \]
   - If \(b = 7\):
     \[
     1 < a < 9 \implies a = 2, 3, 4, 5, 6, 7, 8 \quad (\text{7 values})
     \]

4. Summing up all the possible values of \(a\) for each \(b\):
   \[
   1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
   \]

The final answer is \(\boxed{28}\).