Arrangements with Restrictions

What you need to know

• Block method (glue). To keep $k$ chosen objects together in a row of $n$ different objects, tape them into one block. Arrange the $n-k+1$ things in $(n-k+1)!$ ways, then multiply by $k!$ for the order inside the block. Example: $6$ people in a row with Ann, Ben, and Cal together: $4!\cdot3! = 144$.
• Gap method. To keep $k$ different objects apart (no two side by side), first line up the other $m$ objects. That leaves $m+1$ gaps, counting both ends. Put the $k$ objects into $k$ different gaps: $\binom{m+1}{k}\cdot k!$ ways. Then multiply by the ways to line up the other $m$. Example: $4$ boys and $3$ girls in a row, no two girls together: $4!\cdot\binom53\cdot3! = 24\cdot10\cdot6 = 1440$.
• Round table. $n$ different people sit around a table in $(n-1)!$ ways when turning the table gives the same seating. If flipping it over also gives the same one (beads on a bracelet), it is $\frac{(n-1)!}{2}$, for $n\ge3$. If the seats are numbered, it is $n!$. Example: $5$ people at a round table: $4! = 24$ ways; $5$ beads on a bracelet: $12$.
• Identical objects: $\frac{n!}{a!\,b!\cdots}$, where $a$, $b$, … count each group of identical objects. Example: LEVEL has two L's and two E's, so $\frac{5!}{2!\,2!} = 30$ arrangements.
• Derangements are arrangements with nothing in its own spot. Call the number for $n$ objects $D_n$: $D_2 = 1$, $D_3 = 2$, $D_4 = 9$, $D_5 = 44$. Example: $4$ letters put into $4$ addressed envelopes at random are all wrong with probability $\frac{9}{4!} = \frac{9}{24} = \frac38$.

How AMC 10 tests it

• Problems 8–15: seating in a row or around a table where "A and B must sit together" or "must not sit together."
• Letters of a word with a condition: vowels together, no two vowels side by side, a certain letter not first.
• Problems 15–22: restrictions mixed with symmetry (a round table, necklaces, painted cubes), or "no two side by side" around a circle.
• Often dressed up as a probability: (arrangements that follow the rule) $\div$ (all arrangements).

Standard approaches

1. For "together," glue into a block. For "not together," subtract the "together" count from $n!$. Example: $5$ people, Ann and Ben not side by side: $120-48 = 72$.
2. For "no two of type X side by side," place the others first, count the gaps, then choose gaps.
3. Seat the hardest-to-place object first, then multiply the choices left. On a circle, seat one person first as a fixed reference point; that removes the turned copies of each seating.
4. Do casework on where one special object sits. An end seat has only one neighbor.
5. For $n \le 6$, draw the seats and a small tree of cases.

Worked example

Six students, including Ana and Ben, sit in a row of six chairs. Ana refuses to sit at either end, and Ben must sit next to Ana. How many seatings are possible?

(A) $96$ (B) $144$ (C) $192$ (D) $240$ (E) $288$

Solution. Seat the most limited person first.

Step 1: Ana. She cannot sit at either end, so she takes one of the middle chairs $2$, $3$, $4$, $5$: $4$ choices.

Step 2: Ben. Every middle chair has a chair on each side, so Ben has $2$ choices.

Step 3: everyone else. The other $4$ students fill the $4$ empty chairs in $4! = 24$ ways.

Step 4: multiply. $4\cdot2\cdot24 = 192$.

The answer is $\boxed{\textbf{(C)}\ 192}$.

Pitfalls

• Forgetting the $k!$ for the order inside a block. "Ann, Ben" and "Ben, Ann" are different seatings, so $5$ people with those two together give $4!\cdot2 = 48$, not $24$.
• Using $(n-1)!$ for a circle when the seats are numbered (then it is $n!$), or $n!$ when they are not.
• In a "no two side by side" problem, placing the restricted objects first instead of putting them into the gaps.
• Doing position casework for "A not next to B" and forgetting that an end seat has one neighbor, not two.

Traps that recur

• Counting the labelled colourings and dividing by 8, which fails because three of the twelve classes are left fixed by a reflection; 84 is not even divisible by 8. (2025 AMC 10B #21)
• Merging the two mirror-image spiral shapes into one because they look alike; a rotation can never turn a left-handed spiral into a right-handed one. (2021 AMC Fall 10B #24)
• Trying to place numbers by trial without deriving a structural constraint, or dividing by the wrong symmetry count (24 rotations, not 48 symmetries). (2016 AMC 10A #18)
• Counting the L/R patterns but then assigning the six shoes freely (3! * 3!) instead of noticing that each boundary forces a specific pair to occupy both boundary slots. (2024 AMC 10B #20)