There are 111 StarCraft programmers. The StarCraft team SKT starts with a given set of eleven programmers on it, and at the end of each season, it drops a progamer and adds a programmer (possibly the same one). At the start of the second season, SKT has to field a team of five programmers to play the opening match. How many different lineups of ve players could be fielded if the order of players on the lineup matters?
1. **Initial Setup**: We start with a team of 11 programmers. At the end of the season, one programmer is dropped and another is added. This results in a new team of 11 programmers.

2. **Choosing the Programmer to Drop**: There are 11 options for the programmer to be dropped from the team.

3. **Choosing the Replacement Programmer**: After dropping one programmer, there are 100 remaining programmers (since the dropped programmer can be re-added). Therefore, there are 101 options for the replacement programmer.

4. **Choosing 5 Programmers from the New Team**: From the new team of 11 programmers, we need to choose 5 programmers. The number of ways to choose 5 programmers from 11 is given by the binomial coefficient:
   \[
   \binom{11}{5}
   \]

5. **Ordering the 5 Chosen Programmers**: The order of the 5 chosen programmers matters, so we need to consider all possible permutations of these 5 programmers. The number of permutations of 5 items is given by \(5!\).

6. **Combining All Steps**: The total number of different lineups is the product of all the choices and permutations:
   \[
   11 \cdot 101 \cdot \binom{11}{5} \cdot 5!
   \]

7. **Calculating the Binomial Coefficient and Factorial**:
   \[
   \binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5! \cdot 6!} = 462
   \]
   \[
   5! = 120
   \]

8. **Final Calculation**:
   \[
   11 \cdot 101 \cdot 462 \cdot 120
   \]
   \[
   11 \cdot 101 = 1111
   \]
   \[
   1111 \cdot 462 = 513282
   \]
   \[
   513282 \cdot 120 = 61593840
   \]

Therefore, the total number of different lineups of 5 players is:
\[
\boxed{61593840}
\]