There are 8 events that can be scheduled in a week, then find total number of ways that these 8 events are scheduled on exactly 6 days…[CONT] 
There are 8 events that can be scheduled in a week, then find total number of ways that these 8 events are scheduled on exactly 6 days of a week. (Assume arrangement of multiple events on any particular day as immaterial)

I have been struggling with this question for quite some time now. Here is what I managed to do
The events can be distributed in the following ways
$$1+1+1+1+1+3=8$$
And
$$1+1+1+1+2+2=8$$
Now one of the expressions that apparently gets the answer is
$$\left [ \frac{8!}{3!5!} + \frac{8!}{4! (2!)^2 2!} \right]7!$$
Now the expression inside the bracket seems familiar, basically they are trying to distribute the events on various days, but I could still use an explanation for that. I also have no idea why they multiplied by $7!$
 A: There are two cases.
i) One of the days has $3$ events and each of the remaining $5$ days has an event each.
ii) Two of the days have $2$ events each and rest $4$ days have an event each.
We first make $6$ unordered groups of events. We will assign them to days later.
Number of ways for $i)$ is $ \displaystyle {8 \choose 3} = \frac{8!}{5! \cdot 3!}$.
Number of ways for $ii)$ is $ \ \displaystyle  \frac{1}{2}{8 \choose 2}{6 \choose 2} = \frac{8!}{2^3 \cdot 4!}$
Above, we divide by $2$ as we are making unordered groups at this point.
Now we add $(i)$ and $(ii)$ and order them by multiplying by $7!$ as there are $7$ groups ($6$ with events and one group with no event). If this is not intuitive, we can see it this way - we order $6$ groups with events in $6!$ ways. But we also have $7$ ways to choose the day with no event.
That leads to the answer -
$ \displaystyle \left[ \frac{8!}{5! \cdot 3!} + \frac{8!}{2^3 \cdot 4!} \right] \cdot 7!$
A: Let me give you another approach
Question is obviously disctinct balls into distinct bins , so the exponential generating functions is useful tool for it.
Firstly , lets select which day will not be used ,  we can do it by $C(7,1)=7$ ways.
Now , we have $6$ days and all of them must have at least one event.
Then , the generating function for each day is $$\bigg(x + \frac{x^2}{2!} + \frac{x^3}{3!}\bigg)$$
Then , what we are looking for is that $$\bigg(x + \frac{x^2}{2!} + \frac{x^3}{3!}\bigg)^6 , \bigg[\frac{x^8}{8!} \bigg]$$ or find the coefficient of $x^8$ and multiply it by $8!$ such that https://www.wolframalpha.com/input/?i=expanded+form+of+%28x+%2B+x%5E2+%2F2+%2B+x%5E3+%2F+6%29%5E6
We see that the coefficient of $x^8$ is $\frac{19}{4}$ , then $$\frac{19}{4} \times 8! = 191,520$$
Moreover , we have $C(7,1)=7$ ,so the result is $$C(7,1) \times \frac{19}{4} \times 8! = 1,340,640 $$ which is equal to @Mathlover 's answer
MY ADDITION = If the arrangements of the multiple events on any particular day does matter , then we would think the events as identical objects and the question would turn out to be stars and bars question.
We should select fistly that which day will be omited by $C(7,1)=7$ ways. Then , the $8$ events can be dispersed to $6$ different days by $C(2+6-1,2)=21$ ways where all days has at least one events. (Notice that we thought the events as identical objects).
After that , we should multiply by $8!$ to arrange our events. Notice that when we multpily with $8!$ ,we actually determine that which identical event  is which events .For example , let $$*|*|**|*|**|*$$ be one of the distribution and the leftmost objects in the same gap is the first scheluded event. So , the $8!$ will determine that which star is which event.
As a result , $$7 \times 21 \times 8! =5,927,040$$
