How many ways can 7 persons be arranged at a round table so that 2 particular persons may be together?

7 People: Person 1, Person 2, 3,4,5,6, and 7 at a round table 7 Chairs: 1,2,3,4,5,6,7 Let chairs 1 and 2 be next to each other; Person 1 is at chair 1 and Person 2 is at chair 2 The rest of the people on 3,4,5,6,7 - can sit in 5! ways Next, Person 1 and Person 2 change their positions; Person 1 is sitting at 2 and Person 2 is sitting at 1 The rest of the people on 3,4,5,6,7 - can be rearranged in 5! ways Therefore, total number of ways in which Person 1 and Person 2 are next to each other is: 2*5! = 120*2 = 240 You then have to find in how many ways they are not sitting next to each other; subtract 240 from the total number of possible arrangements; In circular permutation, Total number of arrangements of n people = (n-1)! Here the number of people is 7 Arrangement = (7-1)! = 6!= 720

Number of ways Person 1 and Person 2 are not next to each other = 720-240 = 480

In a circular arrangement we first have to fix the position for the first person, which can be performed in only one way (since every position is considered same if no one is already sitting on any of the seats), also, because there are no mark on positions.

Now, we can also assume that remaining persons are to be seated in a line, because there is a fixed starting and ending point i.e. to the left or right of the first person.

Once we have fixed the position for the first person we can now arrange the remaining $(7-1)$ persons in $(7-1)!= 6!$ ways.