Let  be the set of all sitting arrangements of A, B, C and D in four seats.
As everybody knows,  consists of 4*3*2*1 = 24 permutations.


Let  be the subset of all sitting arrangements of A, B, C and D in four seats, where A is sitting in the seat #1.
Let  be the subset of all sitting arrangements of A, B, C and D in four seats, where B is sitting in the seat #2.
Let  be . . .                                                                  where C is sitting in the seat #3.
Let  be . . .                                                                  where D is sitting in the seat #4.

Then  is, oviously, the union of subsets P +  +  +  + , (1)
where P is our "most desired" subset of those sitting arrangements that are under the problem's question.

It is clear that the cardinality of each subset , ,  and  is 6 = 3*2*1.

Therefore, the first desire is to write 24 = |P| + 6 + 6 + 6 + 6, following (1).
But it would not to be correct.
Why?  - Because the subsets , ,  and  have non-empty intersections that we counted twice.


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Let  be the subset of all sitting arrangements of A, B, C and D in four seats, where A and B are sitting in the seats #1 and #2, respectively.
Let  be the subset of all sitting arrangements of A, B, C and D in four seats, where A and C are sitting in the seats #1 and #3, respectively.
Let , ,  and  be the other four similar subsets in .

Then we can make a correction to formula (1) by writing

 = P +  +  +  +  -  -  -  -  -  -   (2)

in the effort to account for double intersections.
But be aware !!  This formula is still not exactly correct.
Why? - Because the sets , , , ,  and  have triple intersections, which we distracted twice in the formula (2).


Did you just get understanding on how to solve the problem and how to proceed?
If not, then follow me again.


Let  be the subset of all sitting arrangements of A, B, C and D in four seats, where A, B and C are sitting in the seats #1, #2 and #3, respectively.
Let  be the subset of all sitting arrangements of A, B, C and D in four seats, where A, B and D are sitting in the seats #1, #2 and #4, respectively.
Let  and  be the other two similar subsets in .

At last, let  be the subset of all sitting arrangements of A, B, C and D in four seats, where A, B, C and D are sitting in the seats #1, #2, #3 and #4, respectively.
   (As you understand, the set  consists of only ONE element, the identical permitation).

Now, finally, we can write the formula which is ABSOLUTELY true! :

 = P +  +  +  +  -  -  -  -  -  -  +  +  +  +  - .    (3)

I just mentioned above that the cardinality of the 1-letter subsets , ,  and  is 6 = 3*2*1.
The cardinality of the 2-letter subsets , , . . . ,  is 2 = 2*1. Obviously.
The cardinality of the 3-letter subsets , , . . . ,  is 1. Obviously.
The cardinality of the 4-letter subsets  is 1. Obviously.

Therefore, following to the formula (3), we can write its "cardinality" analogue

24 = |P| + (6 + 6 + 6 + 6) - (2 + 2 + 2 + 2 + 2 + 2) + (1 + 1 + 1 + 1) - 1  =  |P| + 4*6 - 2*6 + 4 - 1 = |P| + 24 - 12 + 4 - 1 = |P| + 15.
           ---------------   -----------------------   ---------------
              4 times               6 times                 4 times

Again, our equation is

24 = |P| + 15.

From here, the cardinality of the set P is |P| = 24 - 15 = 9.

Answer.  How many ways are there if Alex does not sit in the first place, Bob does not sit in the second place, 
         Charles does not sit in the third place, and Dan does not sit in the fourth place?   -   9 ways.