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Lets say you have a circular table that seats $n$ people and $b\lt n -1$ identitcal boys. If you were to divide the boys into $k$ teams of size $\geq 1$, how many ways are there to seat the boys so that the teams sit together and have at least 1 empty seat separating them? Consider the seats to be unique (e.g. if we have only 1 boy, there are $n$ possible seating arrangements for him).

Consider the case where we have $n=6$, $b=4$, $k=2$, we can divide the boys into either 2 teams of 2 or a team of 3 and a team of 1. If we label the chairs surrounding the table as $1,2,3,\ldots , 6$, then in the first case, there are 3 unique ways of arranging the boys ((12, 45), (23, 56), (34, 61)) and 6 ways in the second case ((123, 5), (234, 6), $\ldots$ ), leaving 9 total seating arrangements. How can I generalize this? In the case where each team is of size 1, the problem boils down to finding the number of ways of choosing $k$ non-consecutive positions on a ring of size $n$, which has been well-documented (see: Consecutive birthdays probability, e.g.). For bigger teams, the problem seems harder. I suspect I'll first need to calculate the $S(n,k)$, the Stirling number of the second kind to find the total number of possible teams and then find the number of ways of arranging them. This seems to come unwieldy pretty quickly, though.

Anyone have any ideas? Thanks!

Also, can anyone suggest a good reference for this kind of question

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