How many ordered pairs of integers $(a,b)$ satisfy all of the following inequalities? 

\begin{eqnarray*} a^2 + b^2 &<& 16 \\ a^2 + b^2 &<& 8a \\ a^2 + b^2 &<& 8b \end{eqnarray*}
We need to find the number of ordered pairs of integers \((a, b)\) that satisfy the following inequalities:

\[
\begin{aligned}
1. & \quad a^2 + b^2 < 16 \\
2. & \quad a^2 + b^2 < 8a \\
3. & \quad a^2 + b^2 < 8b
\end{aligned}
\]

Let's analyze these inequalities step by step.

1. **First Inequality: \(a^2 + b^2 < 16\)**

   This inequality represents the interior of a circle centered at the origin with radius 4. Therefore, \(a\) and \(b\) must satisfy:

   \[
   -4 < a < 4 \quad \text{and} \quad -4 < b < 4
   \]

   Since \(a\) and \(b\) are integers, the possible values for \(a\) and \(b\) are \(-3, -2, -1, 0, 1, 2, 3\).

2. **Second Inequality: \(a^2 + b^2 < 8a\)**

   Rearrange this inequality:

   \[
   a^2 - 8a + b^2 < 0 \implies (a-4)^2 + b^2 < 16
   \]

   This represents the interior of a circle centered at \((4, 0)\) with radius 4. Since \(a\) and \(b\) are integers, we need to check which of the values from the first inequality satisfy this condition.

3. **Third Inequality: \(a^2 + b^2 < 8b\)**

   Rearrange this inequality:

   \[
   a^2 + b^2 - 8b < 0 \implies a^2 + (b-4)^2 < 16
   \]

   This represents the interior of a circle centered at \((0, 4)\) with radius 4. Again, we need to check which of the values from the first inequality satisfy this condition.

Now, we will check each possible value of \(a\) and \(b\) to see which pairs satisfy all three inequalities.

### Case Analysis

#### Case \(a = 1\)

1. \(1^2 + b^2 < 16 \implies 1 + b^2 < 16 \implies b^2 < 15 \implies -3 < b < 3\)
2. \(1^2 + b^2 < 8 \implies 1 + b^2 < 8 \implies b^2 < 7 \implies -2 < b < 2\)
3. \(1^2 + b^2 < 8b \implies 1 + b^2 < 8b \implies b^2 - 8b + 1 < 0\)

   Solving \(b^2 - 8b + 1 < 0\) using the quadratic formula:

   \[
   b = \frac{8 \pm \sqrt{64 - 4}}{2} = 4 \pm \sqrt{15}
   \]

   Therefore, \(4 - \sqrt{15} < b < 4 + \sqrt{15}\). Since \(\sqrt{15} \approx 3.87\), we get:

   \[
   0.13 < b < 7.87
   \]

   Combining all conditions, \(b\) can be \(1\).

#### Case \(a = 2\)

1. \(2^2 + b^2 < 16 \implies 4 + b^2 < 16 \implies b^2 < 12 \implies -3 < b < 3\)
2. \(2^2 + b^2 < 16 \implies 4 + b^2 < 16 \implies b^2 < 12 \implies -3 < b < 3\)
3. \(2^2 + b^2 < 8b \implies 4 + b^2 < 8b \implies b^2 - 8b + 4 < 0\)

   Solving \(b^2 - 8b + 4 < 0\) using the quadratic formula:

   \[
   b = \frac{8 \pm \sqrt{64 - 16}}{2} = 4 \pm 2\sqrt{3}
   \]

   Therefore, \(4 - 2\sqrt{3} < b < 4 + 2\sqrt{3}\). Since \(2\sqrt{3} \approx 3.46\), we get:

   \[
   0.54 < b < 6.46
   \]

   Combining all conditions, \(b\) can be \(1, 2\).

#### Case \(a = 3\)

1. \(3^2 + b^2 < 16 \implies 9 + b^2 < 16 \implies b^2 < 7 \implies -2 < b < 2\)
2. \(3^2 + b^2 < 24 \implies 9 + b^2 < 24 \implies b^2 < 15 \implies -3 < b < 3\)
3. \(3^2 + b^2 < 8b \implies 9 + b^2 < 8b \implies b^2 - 8b + 9 < 0\)

   Solving \(b^2 - 8b + 9 < 0\) using the quadratic formula:

   \[
   b = \frac{8 \pm \sqrt{64 - 36}}{2} = 4 \pm \sqrt{7}
   \]

   Therefore, \(4 - \sqrt{7} < b < 4 + \sqrt{7}\). Since \(\sqrt{7} \approx 2.65\), we get:

   \[
   1.35 < b < 6.65
   \]

   Combining all conditions, \(b\) can be \(1\).

### Conclusion

Summarizing the valid pairs:

- For \(a = 1\): \(b = 1\)
- For \(a = 2\): \(b = 1, 2\)
- For \(a = 3\): \(b = 1\)

Thus, the total number of solutions is \(1 + 2 + 1 = 4\).

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