Choose a permutation of$ \{1,2, ..., 20\}$ at random. Let $m$ be the amount of numbers in the permutation larger than all numbers before it. Find the expected value of $2^m$.

[i]Proposed by Evan Chang (squareman), USA[/i]
To find the expected value of \(2^m\) where \(m\) is the number of elements in a permutation of \(\{1, 2, \ldots, 20\}\) that are larger than all previous elements, we can use the concept of indicator random variables and linearity of expectation.

1. **Define the Indicator Random Variables:**
   Let \(X_i\) be an indicator random variable that is 1 if the \(i\)-th element in the permutation is larger than all previous elements, and 0 otherwise. Thus, \(m = \sum_{i=1}^{20} X_i\).

2. **Calculate the Probability:**
   The probability that the \(i\)-th element is larger than all previous elements is \(\frac{1}{i}\). This is because any of the first \(i\) elements is equally likely to be the largest among them.

3. **Expected Value of \(X_i\):**
   \[
   \mathbb{E}[X_i] = \frac{1}{i}
   \]

4. **Expected Value of \(m\):**
   Using the linearity of expectation:
   \[
   \mathbb{E}[m] = \mathbb{E}\left[\sum_{i=1}^{20} X_i\right] = \sum_{i=1}^{20} \mathbb{E}[X_i] = \sum_{i=1}^{20} \frac{1}{i}
   \]
   The sum \(\sum_{i=1}^{20} \frac{1}{i}\) is the 20th harmonic number, denoted \(H_{20}\).

5. **Expected Value of \(2^m\):**
   To find the expected value of \(2^m\), we use the fact that \(2^m\) is a function of \(m\). We can use the linearity of expectation again, but this time we need to consider the expectation of \(2^{X_i}\).

6. **Expected Value of \(2^{X_i}\):**
   Since \(X_i\) is an indicator variable:
   \[
   \mathbb{E}[2^{X_i}] = 2^0 \cdot \Pr(X_i = 0) + 2^1 \cdot \Pr(X_i = 1) = 1 \cdot \left(1 - \frac{1}{i}\right) + 2 \cdot \frac{1}{i} = 1 + \frac{1}{i}
   \]

7. **Expected Value of \(2^m\):**
   Using the independence of \(X_i\):
   \[
   \mathbb{E}[2^m] = \mathbb{E}\left[2^{\sum_{i=1}^{20} X_i}\right] = \prod_{i=1}^{20} \mathbb{E}[2^{X_i}] = \prod_{i=1}^{20} \left(1 + \frac{1}{i}\right)
   \]

8. **Simplify the Product:**
   The product \(\prod_{i=1}^{20} \left(1 + \frac{1}{i}\right)\) simplifies to:
   \[
   \prod_{i=1}^{20} \left(1 + \frac{1}{i}\right) = \prod_{i=1}^{20} \frac{i+1}{i} = \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \ldots \cdot \frac{21}{20} = 21
   \]

Therefore, the expected value of \(2^m\) is \(21\).

The final answer is \(\boxed{21}\)