assume that k,n are two positive  integer $k\leq n$count the number of permutation $\{\ 1,\dots ,n\}\ $ st for any $1\leq i,j\leq k$and any positive integer m we have $f^m(i)\neq j$ ($f^m$ meas iterarte function,)
To solve this problem, we need to count the number of permutations of the set $\{1, 2, \ldots, n\}$ such that for any $1 \leq i, j \leq k$ and any positive integer $m$, we have $f^m(i) \neq j$. This means that the first $k$ elements must be in separate cycles in the permutation.

1. **Understanding the Problem**:
   - We are given a set $\{1, 2, \ldots, n\}$.
   - We need to count permutations where the first $k$ elements are in separate cycles.
   - This means that for any $1 \leq i, j \leq k$ and $i \neq j$, $f^m(i) \neq j$ for any positive integer $m$.

2. **Permutations and Cycles**:
   - A permutation can be represented as a product of disjoint cycles.
   - For the first $k$ elements to be in separate cycles, each of these elements must be in a cycle of length 1 (i.e., a fixed point) or in a cycle that does not include any other of the first $k$ elements.

3. **Counting the Permutations**:
   - We need to count the number of ways to arrange the first $k$ elements such that they are in separate cycles.
   - The remaining $n-k$ elements can be arranged in any way.

4. **Deriving the Formula**:
   - The number of ways to arrange the first $k$ elements such that they are in separate cycles is given by the number of derangements of $k$ elements, denoted by $D_k$.
   - A derangement is a permutation where no element appears in its original position.
   - The number of derangements of $k$ elements is given by:
     \[
     D_k = k! \sum_{i=0}^{k} \frac{(-1)^i}{i!}
     \]
   - The remaining $n-k$ elements can be arranged in $(n-k)!$ ways.

5. **Combining the Results**:
   - The total number of permutations is the product of the number of derangements of the first $k$ elements and the number of permutations of the remaining $n-k$ elements:
     \[
     \text{Total permutations} = D_k \cdot (n-k)!
     \]

The final answer is $\boxed{D_k \cdot (n-k)!}$.