In how many ways can people sit around a round table if no two of the people Pierre, Rosa, and Thomas can sit next to each other? (Seating arrangements which are rotations of each other are treated as the same.)
+36 Sorry for the late posting (understatement).
Anyways, you can imagine this question as a group of 5 people and 2 other people - that group of 5 people will always sit next to each other, such that Pierre/Rosa/Thomas will be in the first, third, and fifth position of this group of 5 people. Now, you have 2 positions left, and since there are 4 people left, you can calculate this as \({4 \choose 2}=6\). Also, their are \(3!=6\) ways to have the 3 people moved. Multiplying writing these numbers, you get \(36\). Therefor, you have 36 ways to arrange the group of 5 (whom you can now treat as a single person!). On the other side of the round table, you have 2 people that can be arranged in \(2!=2\) ways, so you have \(36*2=72\), and then multiplying by 3 gives us \(72*3!=72*6=432\). However, since this is a round table, you must divide by 3 since you can rotate the table as
A
B C
or
B
C A
or
C
A B,
Giving us a total of 3 ways you overcounted the original \(432\), so you must divide by 3 to get \(\frac{432}{3}=\boxed{144} \).
(However, you divide by 3 because this question only asks for rotations, but other questions ask for reflections/flips, in which case divide by an additional 2.)
