In this problem, we are asked to determine the number of ways all 11 runners can place in the race, assuming all runners finish.

Step-by-Step Solution:

1. Understanding the Problem: Each runner must be assigned a unique place in the race (1st place, 2nd place, and so on). This means the problem boils down to determining how many ways we can arrange the 11 runners in 11 different positions.

2. Concept: The number of ways to arrange $n$ objects in $n$ distinct positions is given by the formula for permutations: $P(n, n) = n!$ where $n!$ is the factorial of $n$, which means multiplying all integers from $n$ down to 1.

3. Applying the Formula: In this case, $n = 11$. Therefore, the number of ways to arrange 11 runners is: $11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

4. Calculation: Let's calculate $11!$: $11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 39,916,800$

Thus, the number of ways the 11 runners can place in the race is 39,916,800.

Would you like further details on permutations or factorials? Let me know if you'd like to explore more.

Relative Questions:

1. How would the solution change if only the top 3 finishers were considered?
2. What is the factorial of a number, and how does it relate to this problem?
3. How many ways can 5 runners finish in a race if there are 5 entrants?
4. What if two runners tie for a position in the race?
5. How does this problem change if we allow some runners not to finish?

Tip:

When dealing with arrangements where every position must be filled, factorials provide a straightforward way to count all possibilities.