General Solution to the Occupancy Problem with Variably Sized Runs of Adjacent Cells Occupied by Single Balls

Consider the occupancy problem in which (1) there is a row of n cells of which m cells are occupied by one ball each, (2) there are runs of r1, r2, ***, rk adjacent occupied cells with I ri= m that occur in order from one end of the row to the other, and (3) it is desired to find how many different arrangements of occupied and unoccuipied cells there are for the two cases in which there is no restriction on the number of empty cells between runs and in which there is at least one empty cell between each pair of runs. A particular application of this occupancy problem that is easy to grasp involves the seating of theater parties in a row of seats. If parties of size Pi,P2 , IPk with Ek ?i=I pi = m are seated in order in a row of n seats, how many arrangements of occupied and unoccupied seats are there if (1) there need not be a vacant seat between parties, and (2) there is at least one vacant seat between parties? The case in which the parties are all of the same size was solved by an instructive use of recursion by Wiggins [1] including the cases in which each party was composed of two individuals and of m individuals. His model involves the counting of dyadic numbers consisting of l's representing cell partitions and intercalated 0's representing balls occupying cells. The method is related to the one used in Feller [2] to derive the formula for cases in which multiple occupancy of cells is permitted and which considers the number of ways in which the n7-1 partitions between cells or alternatively the balls can be arranged, one to a place, in a total number of places equal to the sum of the n -1 partitions between cells and the balls. However, since at most one ball is permitted to reside in each cell in the problems considered here, the number of cells n figures prominently in the discussion rather than the number of partitions between cells n 1. The intuitive insight that leads to solution of the general case is that the runs of adjacent occupied cells can be considered as occupying single cells if the number of balls is reduced by the difference between the number of balls and the number of runs. The intuitive answer is immediate. In the theater party application let the number of seats be n, the total number of people be m, and the number of parties be k. The number of arrangements of occupied and unoccupied seats is the number of ways in which the n-m empty seats can be chosen from the adjusted number of seats n m + k. The answer is (n7m+k)* Note that the number in each party is not required for solution. Thus, five parties of sizes 100, 5, 5, 5, 5 will generate the same number of arrangements of occupied and unoccupied seats as five parties of sizes 16, 20, 24, 28, and 32 provided the parties are seated in a particular order.