There are 6 people at a party sitting at a round table with 6 seats: A,B,C,D,E AND F. A CANNOT sit next to either D or F. How many ways can the 6 people be seated? A. 720 B. 120 C. 108 D. 84 E. 48 OA is C

sana.noor wrote:There are 6 people at a party sitting at a round table with 6 seats: A,B,C,D,E AND F. A CANNOT sit next to either D or F. How many ways can the 6 people be seated? A. 720 B. 120 C. 108 D. 84 E. 48 OA is C For circular arrangements, we count the number of ways to arrange the REMAINING people RELATIVE to the first person seated. Once A is seated: Number of options for the seat to A's left = 3. (Of the 5 remaining people, anyone but D or F.) Number of options for the seat to A's right = 2. (Of the 4 remaining people, anyone but D or F.) Number of ways to arrange the remaining 3 people = 3! = 6. To combine these options, we multiply: 3*2*6 = 36. The correct answer does not appear among the answer choices. What is the source of this problem?

Another approach - First fix A. Now total possible arrangement - 5! = 120 Arrangements when D and F sit next to = make D and F sit next to A * the rest of available seats with rest of the members = 2! * 3! = 12 120 - 12 = 108 Answer c

hemant_rajput wrote:Another approach - First fix A. Now total possible arrangement - 5! = 120 Arrangements when D and F sit next to = make D and F sit next to A * the rest of available seats with rest of the members = 2! * 3! = 12 120 - 12 = 108 Answer c You have correctly subtracted from the total Bad Case 1: the arrangements in which BOTH D AND F sit next to A. But we must also subtract from the total Bad Case 2: the arrangements in which EITHER D OR F -- but NOT both -- sit next to A. Bad Case 2: Either D or F (but not both) next to A Number of options for the disallowed person sitting next to A = 2. (Either D or F.) Number of seat options for the disallowed person = 2. (To the right or left of A.) Number of options for the OTHER seat next to A = 3. (Anyone but D or F, since only one of them can sit next to A.) Number of ways to arrange the remaining 3 people = 3!. To combine these options, we multiply: 2*2*3*3! = 72. Thus: Total favorable arrangements = total possible arrangements - bad case 1 - bad case 2 = 120-12-72 = 36.