Harry, who is incredibly intellectual, needs to eat carrots $C_1, C_2, C_3$ and solve [i]Daily Challenge[/i] problems $D_1, D_2, D_3$. However, he insists that carrot $C_i$ must be eaten only after solving [i]Daily Challenge[/i] problem $D_i$. In how many satisfactory orders can he complete all six actions?

[i]Proposed by Albert Wang (awang2004)[/i]
1. **Total Permutations**: 
   First, we calculate the total number of permutations of the six actions \(C_1, C_2, C_3, D_1, D_2, D_3\). This is given by:
   \[
   6! = 720
   \]

2. **Pair Constraints**:
   We need to consider the constraints that each carrot \(C_i\) must be eaten only after solving the corresponding Daily Challenge problem \(D_i\). This means for each pair \((D_i, C_i)\), \(D_i\) must come before \(C_i\).

3. **Probability for Each Pair**:
   For each pair \((D_i, C_i)\), there are 2 possible orders: \(D_i\) before \(C_i\) or \(C_i\) before \(D_i\). Only one of these orders satisfies the condition \(D_i\) before \(C_i\). Therefore, the probability that \(D_i\) comes before \(C_i\) is:
   \[
   \frac{1}{2}
   \]

4. **Combined Probability**:
   Since there are three pairs \((D_1, C_1)\), \((D_2, C_2)\), and \((D_3, C_3)\), and each pair must independently satisfy the condition, the combined probability that all three pairs are in the correct order is:
   \[
   \left(\frac{1}{2}\right)^3 = \frac{1}{8}
   \]

5. **Number of Satisfactory Orders**:
   To find the number of satisfactory orders, we multiply the total number of permutations by the probability that all pairs are in the correct order:
   \[
   720 \times \frac{1}{8} = 90
   \]

Conclusion:
\[
\boxed{90}
\]