7 people sit at the round table. In how many ways can we

Question #1: Seven people are to be seated at a round table. Bill and Bob don't want to sit next to each other. How many seating arrangements are possible?

Total # of arrangements around the circular table: \((7-1)!=6!\);
Arrangements when Bill and Bob sit next to each other: 5!*2: consider Bill and Bob as one unit - {Bill, Bob}. Now we have total 6 units: {Bill, Bob}, {1}, {2}, {3}, {4}, {5}. Total # of arrangements of these 6 units around the table is \((6-1)!=5!\) and Bill and Bob within the unit can be arranged in 2 way: {Bill, Bob} and {Bob, Bill};

So, # of arrangements when Bill and Bob don't sit next to each other is: \(6!-2*5!\).

Question #2: Seven people are to be seated at a round table. Bill and Bob don't want to sit opposite to each other. How many seating arrangements are possible?

Now, if we consider "opposite" to mean "directly opposite", meaning that Bill and Bob shouldn't sit at the endpoints of the diameter of the circular table, then total # of arrangements will simply be \((7-1)!=6!\), because if 7 (odd) people are distributed evenly around a circle no two people are directly opposite each other (no diagonal in 7 sided regular polygon passes through the center).

If the question were: SIX people are to be seated at a round table. Bill and Bob don't want to sit opposite to each other. How many seating arrangements are possible?

Total # of arrangements around the circular table: \((6-1)!=5!\);

Now, Bill can have 5 people opposite him. In 1/5 of the cases he'll be opposite Bob (in equal # of arrangements he'll be opposite each of these 5 people), so all but these 1/5 of the cases are acceptable (4/5 of the cases are acceptable). So # of arrangements when Bill and Bob don't sit opposite each other is: \(5!*\frac{4}{5}\).