Why is the probability to have classes every weekday not ${6^5}{25 \choose 2}\left/{30 \choose 7}\right.$? This is 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.)

My wrong answer using naive definition of probability
Total choices are $ ^{30}{C}_7 $. There are buckets with 6 classes each for each of the five days. So we will need to choose one from these buckets. Then from the 25 choices left, choose random 2.
Prob: $\dfrac{(6^5 \cdot ^{25}{C}_2)}{({30}{C}_7)}$
This is greater than 1, so something is obviously wrong. But what exactly?
 A: The problem with your approach is the typical one;: you are counting the same thing more than once. Let $k\in\{1,2,3,4,5\}$ and suppose that the contents of the $k$th bucket consists of $\{a_k,b_k,c_k,d_k,e_k\}$. Then Alice could choose the $a$'s from all buckets, together with $b_1$ and $c_1$. But this single possibility is counted as $3$ different possibilities using your method of counting, since you could group $b_1$ and $c_1$ together, you could group $a_1$ and $c_1$ together, and you could group $a_1$ and $b_1$ together.
A: Ok, suppose first that each class is taught twice a day: once in the morning and once in the afternoon. Then your method for computing the part above the division bar (is it called enumerator in English? I believe so) would be something like this: count all ways she can have one class in the morning each day and, at some point in the week, two classes in the afternoon. (in more detail: the 6^5 are the morning classes and the 25C2 the afternoon classes)
Now we get some info on what is wrong. Suppose that Math and English are both taught at Mondays. You count the event of English on Monday morning and Math in on Monday afternoon and the event of Math on Monday morning and English on Monday afternoon as two separate options. But of course in reality Alice doesn't have that much choice and there is only one way she can have both Math and English. 
