assuming X people sit down around a (round) table, Y people have black shirts and X-Y people have white shirts, what is the probability that the two clusters of shirts are grouped together (i.e., all the black shirts sit together, all white shirts together) around the table?
i've thought about this using standard permutations with circular probability (i.e., (X p Y) / (X-1)!). i've also thought about it sequentially (i.e., someone with a black shirt sits down, so looking at the probability that another black shirt sits down, and another, etc). i get similar results with both methods but do not believe either approach is correct.
thanks in advance for the help and sorry to bore with rudimentary math.
Form a group of all black shirts and add it to the the rest forming a total of $X-Y+1$ objects. Now each of the people inside the group can be shuffled around in $Y!$ . So total number of ways is
$$n=Y!\cdot (X-Y)!$$
Now the total number of ways to seat $N$ people around a table are
$$N=(X-1)!$$
So, the probability is
$$P=\frac{Y!\cdot (X-Y)!}{(X-1)!}$$
This is only valid for the scenario where there is at least $1$ black t-shirt and at least $1$ white t-shirt. For the extreme cases i.e. $Y=X$ and $Y=0$, the probability is simply $1$ as there is no possibility of even forming $2$ groups.