In how many different orders can the characters $P  \ U  \ M  \ \alpha \ C$ be arranged such that the $M$ is to the left of the $\alpha$ and the $\alpha$ is to the left of the $C?$
1. First, we need to determine the total number of ways to arrange the characters \( P, U, M, \alpha, C \). Since there are 5 distinct characters, the total number of permutations is given by:
   \[
   5! = 120
   \]

2. Next, we need to consider the condition that \( M \) is to the left of \( \alpha \) and \( \alpha \) is to the left of \( C \). We can ignore the positions of \( P \) and \( U \) for now and focus on the relative positions of \( M, \alpha, \) and \( C \).

3. The number of ways to arrange \( M, \alpha, \) and \( C \) such that \( M \) is to the left of \( \alpha \) and \( \alpha \) is to the left of \( C \) is just 1, because there is only one way to order them as \( M\alpha C \).

4. The total number of ways to arrange \( M, \alpha, \) and \( C \) without any restrictions is \( 3! = 6 \).

5. Therefore, the probability that \( M, \alpha, \) and \( C \) are in the order \( M\alpha C \) is:
   \[
   \frac{1}{3!} = \frac{1}{6}
   \]

6. Since there are \( 5! \) total arrangements of the characters, and the probability that \( M, \alpha, \) and \( C \) are in the desired order is \( \frac{1}{6} \), the number of valid arrangements is:
   \[
   5! \times \frac{1}{6} = 120 \times \frac{1}{6} = 20
   \]

Conclusion:
\[
\boxed{20}
\]