Seating arrangement at a round table

Question

I am struggling to solve this proplem which goes as follows: we have a round table with 6 seats, on 3 of those seats sits 3 people, with an empty chair between each two, after this 3 more come and sit in the 3 vacant seats, how many diffrent ways can we arrange all 6

Answer 1

So, if I am understanding the question correctly... we have three early people... $A,B,C$ and then three late people $X,Y,Z$. The early people will not sit next to one another, and the table is "round" meaning that one seating arrangement and another seating arrangement are considered the "same" if the first is just a rotation of the other.

Without loss of generality, let $A$ sit at the northernmost seat available. It doesn't matter which seat $A$ sits in specifically, we will just use $A$ as a reference point from here out in order to avoid ambiguity.

Now, let $B$ and $C$ pick their seats at the table in relation to $A$, again respecting the rules that they can't sit in chairs adjacent to each other. There are $2$ ways this can happen.

Once they are situated, let $X,Y,Z$ arrive and choose which of the three remaining chairs $X$ sits at. Then choose which chair $Y$ sits at, and lastly $Z$ will take the final seat.

$$2\times 3! = 12$$

Answer 2

For the first condition, you need to arrage 3 people on 6 seats so that there is an empty seat between two people. If you do not count repeated cases, there is only two cases that can be arranged. For the second condition, 3 more people came and sit in the 3 vacant seats, is the same thing as arranging 3 people with 3 vacant space so the number of ways is 3!. So there can be 2 inital cases, with 3! more further cases per initial case, So $2(3!)=2(6)=12$ cases or ways.