$\textbf{Problem 2.}$ The list of all the numbers of $5$ different numbers that are formed with the digits $1,2,3,4$ and $5$ is made. 
In this list the numbers are ordered from least to greatest. Find the number that occupies the $100th$ position in the list.
1. **Determine the total number of permutations:**
   The problem involves forming numbers using the digits \(1, 2, 3, 4, 5\). Since all digits are used and each digit is unique, the total number of permutations is given by \(5!\):
   \[
   5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
   \]
   Therefore, there are 120 different numbers that can be formed.

2. **Group the permutations by the first digit:**
   Each digit can be the first digit, and for each choice of the first digit, there are \(4!\) permutations of the remaining four digits:
   \[
   4! = 4 \times 3 \times 2 \times 1 = 24
   \]
   Thus, there are 24 numbers starting with each digit.

3. **Identify the range of numbers:**
   - Numbers starting with \(1\): Positions 1 to 24
   - Numbers starting with \(2\): Positions 25 to 48
   - Numbers starting with \(3\): Positions 49 to 72
   - Numbers starting with \(4\): Positions 73 to 96
   - Numbers starting with \(5\): Positions 97 to 120

4. **Locate the 100th number:**
   Since the 100th number falls within the range of numbers starting with \(5\) (positions 97 to 120), we need to find the 4th number in the sequence starting with \(5\):
   \[
   100 - 96 = 4
   \]
   So, we need the 4th permutation of the digits \(1, 2, 3, 4\) following the digit \(5\).

5. **List the permutations of \(1, 2, 3, 4\):**
   The permutations of \(1, 2, 3, 4\) in lexicographic order are:
   \[
   1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321
   \]
   The 4th permutation is \(1342\).

6. **Combine the first digit with the 4th permutation:**
   The 100th number in the list is:
   \[
   5 \text{ followed by } 1342 \implies 51342
   \]

The final answer is \(\boxed{51342}\)