The integers $ 1,2,...,n$ are placed in order so that each value is either strictly bigger than all the preceding values or is strictly smaller than all preceding values. In how many ways can this be done?
1. Consider the sequence of integers \(1, 2, \ldots, n\). We need to arrange these integers such that each value is either strictly bigger than all the preceding values or strictly smaller than all the preceding values.

2. Let the permutation start with the integer \(k\). This means \(k\) is the first element in the sequence. 

3. The integers \(1, 2, \ldots, k-1\) must appear in the sequence in strictly decreasing order after \(k\). This is because if any of these integers appeared in any other order, it would violate the condition that each value is either strictly bigger or strictly smaller than all preceding values.

4. Similarly, the integers \(k+1, k+2, \ldots, n\) must appear in strictly increasing order after \(k\). This ensures that the condition is maintained for all values greater than \(k\).

5. For each \(k\), there are \(\binom{n-1}{k-1}\) ways to choose \(k-1\) slots for the numbers smaller than \(k\). This is because we are choosing \(k-1\) positions out of \(n-1\) available positions (excluding the position of \(k\)).

6. Summing across all values of \(k\) from \(1\) to \(n\), we get the total number of ways to arrange the sequence. This sum can be expressed as:
   \[
   \sum_{k=1}^{n} \binom{n-1}{k-1}
   \]

7. Using the binomial theorem, we know that:
   \[
   \sum_{k=0}^{n-1} \binom{n-1}{k} = 2^{n-1}
   \]
   Since \(\binom{n-1}{k-1}\) is just a shift of the binomial coefficients, the sum from \(k=1\) to \(n\) is equivalent to the sum from \(k=0\) to \(n-1\).

8. Therefore, the total number of ways to arrange the sequence is:
   \[
   \sum_{k=1}^{n} \binom{n-1}{k-1} = 2^{n-1}
   \]

Conclusion:
\(\boxed{2^{n-1}}\)