Order $25$ people in $4$ circles, $3$ circles with $6$ people and $1$ circle with $7$ people - the order between the circles isn't important I am learning to my discrete math test and I tried to answer this question.
I know how to count the options with importance of the order between the circles, using the multiplication principle.
I count the options to put $6$ of $25$ people in the first circle - $p(25,6)$ and options to order them in the circle which is $5!$
Then count the options to put $19$ of $25$ people in the second circle - $p(19,6)$ and the options to order them in the circle which is $5!$
Then count the options to put $13$ of $25$ people in the second circle - $p(13,6)$ and the options to order them in the circle which is $5!$
Then count the options to put $7$ of $7$ people in the second circle - $1$ and the options to order them in the circle which is $6!$
Then I multiply all of them and I get that the number of options to order the people in the circles is: $(25, 6) \times (19, 6) \times (13, 6) \times (5!)^3 \times 6!$
Now I need to use diverse principle - I know that for every group of circles there are $4!$ different ways to order the circles, so if I am dividing the count with importance of the order by $4!$ I get the answer.
The problem is that in the solution they say that I need to diverse it by $3!$ because I already know that the circle with $7$ people will be last in the order and i am having struggles to understand the idea.
The question is as I write in the header, so I hope I was clear enough that people can help me get the idea in the solution.
 A: Rather than dividing by symmetry, perhaps a more direct argument will make more sense.
First, let us choose the seven people who sit at the seven-person-table.  There is no ambiguity as to which table is the seven-person-table, it is the only table with seven chairs.  We choose those seven people in $\binom{25}{7}$ ways.
Once having chosen those people, we look at them and one of those people will be youngest.  Let them sit at the seven-person-table in whatever chair they like, it doesn't matter to us since we aren't keeping track of that information.  From there, take the remaining six people and arrange them around the remaining seats of the table.  We do keep track of this information since we care about how the people are seated with relation to each other (though we don't care how they are seated with relation to the door to the room).  This can be done in $6!$ ways.
From here, and this is where my proposed solution differs from yours, among the remaining people there will be some youngest person and that person will eventually be seated somewhere.  Go ahead and let them choose somewhere to sit.  It doesn't bother us which of the remaining tables they sit at, they all look the same to us.  It doesn't matter which of the seats they sit at either since we only care how people are seated relative to one another, not how the table is situated around the room or where the chair is relative to the entrance etc...
Now that the youngest has chosen their seat, choose five more people to sit at the table with them and then choose how they are arranged around the table relative to that youngest person.  This can be done in $\binom{17}{5}\cdot 5!$ ways.
Now, among the remaining $12$ people, again, there will be a youngest remaining.  Let them sit in one of the remaining available seats at one of the remaining available tables, again it matters not which.  Choose the $5$ other people to sit with them and arrange around the table in $\binom{11}{5}\cdot 5!$ ways.  Do this one final time for the final set of people in $\binom{5}{5}5!$ ways.
This gives a grand total of:
$$\binom{25}{6}6!\cdot \binom{17}{5}5!\cdot \binom{11}{5}5!\cdot \binom{5}{5}5!$$

As for your division by symmetry approach and dividing by $3!$ instead of by $4!$, the point being that when choosing which table they sat at, as alluded to in the teacher's solution we can correctly distinguish which of the tables is the seven-person-table by the fact that it is the only table with seven people at it... this even if the labels of the table were hidden from us.  The three six-person-tables however if left unlabeled we would be unable to distinguish so our initial approach to counting which had these tables labeled as the "first" "second" and "third" had counted each discernable outcome six times, once for each different arrangement of the labels on the tables.
A: Why do you need to divide by $3!$ and not by $4!$?
Note all the circles of size $6$ can be interchanged with one another, while the circle of size $7$ is "special" - cannot be swapped with any other circle. Your solution before the last division assumed you knew which of the circles of "size" $6$ was "first", which was "second" and which was "third". Now, as we don't care about the order of those three circles, we divide by the number of permutations of those circles, i.e. $3!$.
To illustrate further: in a hypothetical similar problem, where $26$ people would be split into two circles of size $7$ and two circles of size $6$, you would, as the last step, need to divide by $2!\cdot 2! = 4$, because the two circles of size $7$ can be swapped, and two circles of size $6$ can be swapped, but a circle of size $6$ cannot be swapped for a circle of size $7$.
