Why isn't the probability that Alice will have classes every weekday $\dfrac{6^5 \times 25 \times 24}{30\choose 7}$?? Consider this exercise  from Harvard Stat 110 Strategic Practice 2, Fall 2011, and Blitzstein's Introduction to Probability (2019 2 ed) Ch 1, Exercise 54, p 51.

Alice attends a small college in which each class meets only once a week. She is deciding between 30 non-overlapping classes. There are 6 classes to choose from for each day of the week, Monday through Friday. Trusting in the benevolence of randomness, Alice decides to register for 7 randomly selected classes out of the 30, with all choices equally likely. What is the probability that she will have classes every day, Monday through Friday? (This problem can be done either directly using the naive definition of probability, or using inclusion-exclusion.)"

Here's the solution manual's Direct Method (Not Inclusion-Exclusion).

There are two general ways that Alice can have class every day: either she has 2 days with 2 classes and 3 days with 1 class, or she has 1
day with 3 classes, and has 1 class on each of the other 4 days. The number of possibilities for the former is:


${5 \choose 2}{6 \choose 2}^26^3$ (choose the 2 days when she has 2 classes, and then select 2 classes on those days and 1 class for the other days).


The number of possibilities for the latter is:


${5 \choose 1}{6 \choose 3}6^4$


So the probability is: $\frac{{5 \choose 2}{6 \choose 2}^26^3 + {5 \choose 1}{6 \choose 3}6^4}{30\choose 7}$ which is close to 30.2%.

Please consider my wrong answer, and let me know what's wrong?
My answer: $\frac{6^5 \times 25 \times 24}{30\choose 7}$
Reasoning: For every one of the 5 days of the week, Alice has 6 choices.Thus $6^5$.
But she still has to choose 2 more classes, out of 25 classes remaining. Thus $25 \times 24$.
The denominator is the same in both cases, so no need for further explanation. Choose 7 classes out of 30.
Needless to say, my reasoning does not give the desired result. I'm concerned about the way I approach such problems. My reasoning is flawed, and it is affecting more than this exercise.

If Alice has more than one class in a day, your method counts it more than once. You count each day in which two classes meet twice, once for each of the two ways you could have designated one of those two classes that meet that day as the class that meets that day.  Similarly, you count each day in which three classes meet three times, once for each of the three ways you could have designated one of the classes that meets that day as the class that meets that day.
Also, you failed to account for the fact that there are $\binom{5}{1} = 5$ ways to have one day on which three classes meet and $\binom{5}{2} = 10$ ways to have two days on which two classes meet.  These are mutually exclusive cases, which is why you need to add.
Finally, note that $25 \cdot 24$ is an ordered selection, but the order in which the classes are chosen does not matter.
