Why Restricted Arrangements Need Care
Counting an unrestricted arrangement is often a direct application of factorials. Restrictions change the task: a required position removes choices, adjacency links objects into a unit, and separation forces objects into different gaps. The main challenge is to represent the restriction correctly before multiplying.
This tutorial treats objects as distinct and positions as ordered. For example, arranging seven different folders on a shelf is not the same as arranging seven identical folders. The methods also apply to people in seats, labeled tasks in a schedule, or distinct components in a sequence. Each scenario below is hypothetical.
A useful starting point is to name the positions and translate each condition literally. “A is in position 3” fixes A in one slot. “A and B are adjacent” means they occupy consecutive slots, in either order. “A, B, and C are never adjacent to one another” means no two of those three occupy consecutive slots. These are different conditions and require different counting models.
A Reliable Counting Plan
Decide whether each object has a unique identity and whether each position matters. Factorials count orderings of distinct objects in distinct positions.
Fix a position, split into separate cases, combine adjacent objects into a block, or reserve gaps to enforce separation.
When using cases, make them disjoint. When using a block, count its internal order. When using gaps, count which gaps are selected and how the designated objects are ordered.
Confirm that every arrangement satisfying the restriction is represented and that no arrangement is counted more than once.
Required Positions and Disjoint Cases
A fixed position is handled by placing the required object first and then arranging the objects that remain. If an object is allowed in several positions, split the count according to its position. The cases must not overlap: an object cannot occupy two different positions in the same arrangement.
Worked Example: Two Allowed Positions and One Fixed Position
Eight distinct volunteers are assigned to eight positions in a display row. Volunteer M must be in position 4, and volunteer L must be in position 1 or position 8. How many assignments are possible? Use disjoint cases, and identify the common mistake avoided.
Method. Split according to L’s position. The cases “L is in position 1” and “L is in position 8” are disjoint, and together they include every permitted placement of L.
Case 1: L is in position 1. M is in position 4. Those two positions are occupied, leaving six distinct volunteers for the other six positions. They can be arranged in \(6!=720\) ways.
Case 2: L is in position 8. M is still in position 4. Again, six distinct volunteers fill six remaining positions in \(6!=720\) ways.
Add the disjoint case counts:
Check. Every valid assignment puts L at exactly one of the two permitted ends, and M is fixed in both cases. The answer is 1,440 assignments. The common mistake is to treat the two allowed positions for L as if they were both occupied; the cases instead represent alternatives.
Adjacency: Treat a Pair as a Block
To require two specified objects to be adjacent, temporarily combine them into one block. Arrange that block with the other objects, then account for the orders inside the block. For two distinct objects, there are two internal orders: AB and BA. The block method works because an adjacent pair occupies consecutive positions and can therefore be represented by one unit in the larger ordering.
An additional position restriction must still be checked in the original arrangement. In particular, if a person outside the block is forbidden from the first position, the block can occupy the first position without violating that condition. The person is at the start of the row only when that person’s unit is first.
Worked Example: An Adjacent Pair and a Forbidden First Position
Seven distinct identification badges are displayed in a row. Badges D and E must be adjacent, and badge F must not be in position 1. How many displays satisfy both restrictions? Use a block and count directly.
Method. Treat D and E as one block. There are then six units to order: the DE block and the five other badges, including F. F is not in the first position precisely when its unit is not first in this six-unit ordering.
Choose the position of F’s unit among positions 2 through 6. There are \(5\) choices. The other five units, including the DE block, can then be ordered in \(5!=120\) ways. Finally, D and E can appear inside the block in \(2!=2\) orders.
Check. Each counted display has the D–E pair consecutive, F is not first, and both internal orders of the pair are included. Conversely, every permitted display gives one choice of F’s unit position, one ordering of the other units, and one internal pair order. The answer is 1,200. A common mistake is to count the block’s placement but forget its two internal orders, or to exclude a first-position block even though that does not put F first.
Separation: Use Gaps
When designated objects must not be adjacent to one another, first arrange the objects that are not restricted. This creates gaps before, between, and after them. Put at most one designated object in each gap. That guarantees a separating object between every pair of designated objects.
To justify the result, fix an ordering of the \(m\) other objects. Every permitted arrangement of the designated objects determines exactly which gaps they occupy and which designated object goes in each chosen gap. Conversely, selecting distinct gaps and assigning the designated objects to them yields an arrangement with no two designated objects adjacent. Therefore, for each fixed ordering of the other objects, the number of placements is \(\binom{m+1}{r}r!\). There are \(m!\) possible orderings of those other objects, so the total is \(m!\binom{m+1}{r}r!\).
Worked Example: Three Speakers Must Be Separated
Seven distinct speakers are assigned to a speaking order. Three named speakers, A, B, and C, must not be adjacent to one another; the other four speakers may appear anywhere. How many speaking orders are permitted? Use gaps, and identify the mistake the method prevents.
Method. First arrange the four speakers who are not A, B, or C. They can be ordered in \(4!=24\) ways. For any such order, there are five gaps: before the first speaker, three between consecutive speakers, and after the last speaker.
Choose three different gaps for A, B, and C. There are \(\binom{5}{3}=10\) choices. Arrange A, B, and C among the chosen gaps in \(3!=6\) ways. Thus:
Check. No gap receives more than one of A, B, and C, so at least one of the other speakers separates every pair of designated speakers. Every valid order has a unique ordering of the other four speakers, a unique set of occupied gaps, and a unique order of A, B, and C in those gaps. The answer is 1,440. The common mistake is to count only the choice of three gaps and omit the \(3!\) ways to assign the named speakers to them.
Combining Restrictions Without Losing Track
Several restrictions can be handled in one count, but it is important to check how they interact. A block reduces the number of units being ordered. A fixed-position condition removes a position from the remaining choices. A gap condition controls where designated objects can be placed. Apply each step to the objects and positions still available, rather than multiplying separate counts that may describe the same choices or incompatible arrangements.
For example, if two objects must be adjacent while a third must occupy a specified position, first check whether the specified position could belong to the block. If it does, the block may have more possible placements than if it does not. Similarly, if one object has two allowed positions and another has a fixed position, verify that the positions do not coincide in a way that would require two distinct objects in one slot. If they do, separate the valid cases explicitly.
The methods here are useful whenever the event of interest is defined by an arrangement condition. Once the valid arrangements have been counted, a probability can be found in a finite model if outcomes are equally likely. The counting step itself does not establish that outcomes are equally likely; that is a separate modeling assumption.
Common Mistakes
- Forgetting internal orders in a block. A two-object block has two orders. A block of \(r\) distinct objects has \(r!\) internal orders.
- Using overlapping cases. “A is in position 1” and “A is in position 8” are disjoint. Cases such as “A is in an odd position” and “A is before B” may overlap, so they cannot automatically be added.
- Confusing a block position with an actual position. A block is one unit for ordering purposes but occupies multiple consecutive positions in the original row.
- Leaving out assignments to gaps. Choosing gaps selects locations, but distinct designated objects must also be assigned to those gaps.
- Applying a method to the wrong restriction. A block enforces adjacency. Separate gaps enforce nonadjacency. Reversing these models gives the wrong set of arrangements.
Check Your Understanding
For each problem, assume all named objects are distinct and all positions are ordered.
- Six technicians are assigned to six different stations. Technician A must be at station 2, and technician B must be at station 1 or station 6. How many assignments are possible? (Answer: 48)
- Eight distinct folders are placed in a row. Folders K and L must be adjacent, and folder C must be in position 1. How many arrangements are possible? (Answer: 1,440)
- Seven distinct speakers are ordered in a row. Three named speakers, A, B, and C, must not be adjacent to one another. How many orders are possible? (Answer: 1,440)
- Seven distinct labels are placed in a row. P must be in position 1 or position 7, and Q must not be adjacent to P. How many arrangements satisfy both restrictions? (Answer: 1,200)
- Nine distinct books are arranged on a shelf. Three named books must occupy the first three positions in any order. Two other specified books must be adjacent to each other among positions 4 through 9. How many arrangements are possible? (Answer: 1,440)