To solve this problem, we need to calculate the number of ways to select and arrange 5 classes out of 11, where the order matters, and no class is repeated.

This is a permutation problem, where the number of permutations of $r$ objects taken from a set of $n$ distinct objects is given by:

$P(n, r) = \frac{n!}{(n-r)!}$

In this case:

• $n = 11$ (the total number of classes)
• $r = 5$ (the number of classes to choose)

So, we need to calculate $P(11, 5)$:

$P(11, 5) = \frac{11!}{(11-5)!} = \frac{11!}{6!}$

Now, calculate the factorials:

$11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6!$

Notice that $6!$ in the numerator and denominator cancels out, so:

$P(11, 5) = 11 \times 10 \times 9 \times 8 \times 7$

Now, calculate the product:

$P(11, 5) = 11 \times 10 \times 9 \times 8 \times 7 = 55440$

So, there are 55,440 ways for the nursing students to schedule their 5 classes in order.

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:

1. How many ways can the nursing students choose 4 classes out of 11 in order?
2. If the students need to choose 5 classes out of 11 but the order doesn't matter, how many ways can they do it?
3. What if the students have to schedule their classes without overlapping time slots—how does this affect the number of possible schedules?
4. How many ways can the students schedule their classes if 2 out of the 11 classes are mandatory and must be taken first?
5. How many different schedules are possible if students can repeat classes, i.e., schedule the same class more than once?

Tip: Permutations focus on arranging a subset in order, whereas combinations deal with selecting a subset without considering the order.