Why isn't the probability that Alice will have classes every weekday $\dfrac{5\times \binom{5}{2}}{30\choose 7}$? 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 thinking was to first assign one class to each of the $5$ days, $6^5$ ways of doing this. Then multiply this with the probability of selecting remain $2$ classes such that
$a)$ either they both are on the same day, or
$b)$ on two different days.
Probability for $(a)= 5\times \binom{5}{2}$. Prob. of $(b)=\binom{5}{2}\times 5$. This gives total no. of ways to assign classes as required. Then divide this by $\binom{30}{7}$ $($total no of ways to assign classes randomly$)$. But this gives a probability greater than $1$ . Where is my thinking wrong? All classes are equally likely, and I don't think the process of choosing follows order of days.
 A: There are two options:


*

*Select 2 days with two classes and 3 days with one class
$$|A|=\binom{5}{2}\binom{6}{2}^2\binom{6}{1}^3=486000$$

*Select 1 day with three classes and 4 days with one class
$$|B|=\binom{5}{1}\binom{6}{3}\binom{6}{1}^4=129600$$


Of course $A\cap B = \emptyset$ and the full scope of selection $\Omega$ contains $|\Omega|=\binom{30}{7}=2035800$ elements. Thus
$$P = \frac{|A|+|B|}{|\Omega|} = \frac{615600}{2035800} \approx 0.3$$
Your attempt
You're missing some details. 
For selecting two days with additional classes:


*

*Select one class for each day ($6^5$)

*Select two days ($\binom{5}{2}$)

*Select additional classes for these days ($\binom{5}{1}^2$)

*Avoid counting situations multiple times by dividing it by the number of possible permutations of selection ($(2!)^2$)


So the number of distinct events in that case is
$$|A|=\frac{6^5\binom{5}{2}\binom{5}{1}^2}{(2!)^2}$$
Analogically we can calculate number of distinct events in case, where we select only one day with three classes (divided by number of possible selections of classes in day with three classes as the first selected ($3$)):
$$|B|=\frac{6^5\binom{5}{1}\binom{5}{2}}{3}$$
A: There are $\binom{24}{7}$ selections that are avoiding one of the five days.
There are $\binom{18}{7}$ selections that are avoiding two of the five days.
There are $\binom{12}{7}$ selections that are avoiding three of the five days.
The number of selections that are avoiding one or more days is, by inclusion-exclusion :
$bad = 5\binom{24}{7} - 10 \binom{18}{7} + 10\binom{12}{7} - 0 + 0 = 1420200  $
$\frac {total - bad} {total} = \frac{615600}{2035800} \approx 0.30$ 
A: p 8 from the Selected Solutions PDF to Blitzstein's Introduction to Probability (2019 2 ed) Ch 1, Exercise 54, p 51.

This can be done by two methods.
Direct Method
Inclusion-Exclusion Method
Lets Starts with
Direct Method:- 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 $\dbinom{5}{2}\dbinom{6}{2}6^3$ $($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 $\dbinom{5}{1}\dbinom{6}{3}6^4$. So the probability is
$$\dfrac{\dbinom{5}{2}\dbinom{6}{2}6^3+\dbinom{5}{1}\dbinom{6}{3}6^4}{\dbinom{30}{7}}=\dfrac{114}{377}\approx0.302$$
Inclusion-Exclusion Method- We will use inclusion-exclusion to find the probability of the complement, which is the event that she
has at least one day with no classes.
Let say $B_i=A^c_i.$ Then $$P(B_1\cup B_2....\cup B_5)=\sum _{ i }^{ 
> }{ P(B_ i)- } \sum _{ i<j }^{  }{ P(B_ i)-\sum _{ i<j }^{  }{ P(B_
> i\cap B_ j)+\sum _{ i<j<k }^{  }{ P(B_ i\cap B_ j\cap B_ k) }  }  } $$
Note that terms with the intersection of $4$ or more $B_i$’s are not
needed since Alice must have classes on at least $2$ days. Then we
have  $$P(B_1)=\dfrac{\dbinom{24}{7}}{\dbinom{30}{7}},P(B_1\cap
> B_2)=\dfrac{\dbinom{18}{7}}{\dbinom{30}{7}},P(B_1\cap B_2\cap
> B_3)=\dfrac{\dbinom{12}{7}}{\dbinom{30}{7}}$$and similarly for the
other intersections. So$$P(B_1\cup B_2....\cup
> B_5)=5\dfrac{\dbinom{24}{7}}{\dbinom{30}{7}}-\dbinom{5}{2}\dfrac{\dbinom{18}{7}}{\dbinom{30}{7}}+\dbinom{5}{3}\dfrac{\dbinom{12}{7}}{\dbinom{30}{7}}=\dfrac{263}{377}$$
Therefor, $$P(A_1\cap A_2\cap A_3\cap A_4\cap
> A_5)=\dfrac{114}{377}\approx0.302$$

