I asked a similar question here which is as follows:
Four (identical) persons enter a train (section A has 4 seats, section B has 8 seats). What is the probability that exactly (not more or less) 3 persons sit in section A? (all seats have the same probability)
This problem is known as one of the urn problems and follows the hypergeometric distribution (Wikipedia). The probability can be calculated this way:
$$\frac{\binom{R}{r}\binom{N-R}{n-r}}{\binom{N}{n}}=\frac{\binom{4}{3}\binom{8}{1}}{\binom{12}{4}}$$
I now would like to calculate the probability that exactly (not more or less) 3 persons sit in section A IF MAXIMUM 2 PEOPLE CAN SIT ON ONE SEAT AT THE SAME TIME?
What I could find out so far is that if one selects r of the n seats and allowing that a seat can be chosen again, the number of possibilities can be calculated by
$$\binom{n+r-1}{r}$$
Unfortunately one can't specify a maximum.
It's really the same as having twice as many seats in each section with the original maximum of one person per seat. So the answer would be:
$$\frac{\binom{2R}{r}\binom{2N-2R}{n-r}}{\binom{2N}{n}}=\frac{\binom{8}{3}\binom{16}{1}}{\binom{24}{4}}$$
where, as before, \begin{eqnarray*} N &=& \mbox{total seats in train} \\ R &=& \mbox{total seats in Section A} \\ n &=& \mbox{number of persons} \\ r &=& \mbox{number of seats chosen in Section A.} \end{eqnarray*}