Combinatorics: Block Method 2, Round Table

What is the probability that two specific people always sit together when 10 people are seated around a circular table?

Answer

Because seating is rotationally invariant, we fix a reference person (w.l.o.g. assumed not to be one of the two specified individuals). Then, there are $$9!$$ possible arrangements for the remaining 9 people.

Now, to count favorable outcomes we treat the two specified people as a single block. With the reference fixed, there are $$10 - 1 - 2 = 7$$ other individuals whose order can be arranged in $$7!$$ ways. In the circle of these 7 people, there are $$8$$ gaps (think about it) where we can insert the block. This ensures the condition is fulfilled. Finally, the two people within the block can be arranged in 2 ways.

Thus, the number of favorable arrangements is:

2 \times 8 \times 7!.

Therefore, the probability that the two specified people sit together is:

\frac{2 \times 8 \times 7!}{9!} =\frac{2}{9}.

\boxed{\frac{2}{9}}