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Arrangements with Restrictions

Permutations with constraints, adjacency, circular arrangements, forbidden positions

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Arrangements with RestrictionsAMC 10 Step by Step

Name: Date: October 28, 2026

20 problems, easiest first

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

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: $120-48 = 72$ .
For "no two of type X side by side," place the others first, count the gaps, then choose gaps.
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.
Do casework on where one special object sits. An end seat has only one neighbor.
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)

1.2018 AMC 10A #4
How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
1. A) $3$
2. B) $6$
3. C) $12$
4. D) $18$
5. E) $24$
2.2014 AMC 10A #4
Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?
1. A) $2$
2. B) $3$
3. C) $4$
4. D) $5$
5. E) $6$
3.2022 AMC 10A #9
A rectangle is partitioned into $5$ regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?
1. A) $120$
2. B) $270$
3. C) $360$
4. D) $540$
5. E) $720$
4.2015 AMC 10A #10
How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either $ab$ or $ba$ .
1. A) $0$
2. B) $1$
3. C) $2$
4. D) $3$
5. E) $4$
5.2012 AMC 10B #11
A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
1. A) $729$
2. B) $972$
3. C) $1024$
4. D) $2187$
5. E) $2304$
6.2025 AMC 10B #16
A circle has been divided into $6$ sectors of different sizes. Then $2$ of the sectors are painted red, $2$ painted green, and $2$ painted blue so that no two neighboring sectors are painted the same color. One such coloring is shown below.

How many different colorings are possible?
1. A) $12$
2. B) $16$
3. C) $18$
4. D) $24$
5. E) $28$
7.2024 AMC 10B #17
In a race among $5$ snails, there is at most one tie, but that tie can involve any number of snails. For example, the result might be that Dazzler is first; Abby, Cyrus, and Elroy are tied for second; and Bruna is fifth. How many different results of the race are possible?
1. A) $180$
2. B) $361$
3. C) $420$
4. D) $431$
5. E) $720$
8.2022 AMC 10A #14
How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?
1. A) $108$
2. B) $120$
3. C) $126$
4. D) $132$
5. E) $144$
9.2021 AMC Fall 10A #18
A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field?
1. A) $12$
2. B) $64$
3. C) $84$
4. D) $90$
5. E) $144$
10.2017 AMC 10A #19
Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of 5 chairs under these conditions?
1. A) $12$
2. B) $16$
3. C) $28$
4. D) $32$
5. E) $40$
11.2000 AMC 10 #13
There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?
1. A) $0$
2. B) $1$
3. C) $5!\cdot 4!\cdot 3!\cdot 2!\cdot 1!$
4. D) $\frac{15!}{5!\cdot 4!\cdot 3!\cdot 2!\cdot 1!}$
5. E) $15!$
12.2024 AMC 10B #20
Three different pairs of shoes are placed in a row so that no left shoe is next to a right shoe from a different pair. In how many ways can these six shoes be lined up?
1. A) $60$
2. B) $72$
3. C) $90$
4. D) $108$
5. E) $120$
13.2021 AMC Fall 10A #24
Each of the $12$ edges of a cube is labeled $0$ or $1$ . Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$ ?
1. A) $8$
2. B) $10$
3. C) $12$
4. D) $16$
5. E) $20$
14.2021 AMC 10A #20
In how many ways can the sequence $1, 2, 3, 4, 5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
1. A) $10$
2. B) $18$
3. C) $24$
4. D) $32$
5. E) $44$
15.2018 AMC 10B #18
Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
1. A) $60$
2. B) $72$
3. C) $92$
4. D) $96$
5. E) $120$
16.2017 AMC 10B #18
In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
1. A) $6$
2. B) $8$
3. C) $9$
4. D) $12$
5. E) $15$
17.2014 AMC 10B #24
The numbers 1, 2, 3, 4, 5 are to be arranged in a circle. An arrangement is bad if it is not true that for every $n$ from $1$ to $15$ one can find a subset of the numbers that appear consecutively on the circle that sum to $n$ . Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?
1. A) $1$
2. B) $2$
3. C) $3$
4. D) $4$
5. E) $5$
18.2013 AMC 10B #22
The regular octagon $ABCDEFGH$ has its center at $J$ . Each of the vertices and the center are to be associated with one of the digits $1$ through $9$ , with each digit used once, in such a way that the sums of the numbers on the lines $AJE$ , $BJF$ , $CJG$ , and $DJH$ are all equal. In how many ways can this be done?
1. A) $384$
2. B) $576$
3. C) $1152$
4. D) $1680$
5. E) $3456$
19.2012 AMC 10B #22
Let $(a_1,a_2, \dots ,a_{10})$ be a list of the first 10 positive integers such that for each $2 \le i \le 10$ either $a_i+1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?
1. A) $120$
2. B) $512$
3. C) $1024$
4. D) $181,440$
5. E) $362,880$
20.2011 AMC 10A #22
Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
1. A) $2520$
2. B) $2880$
3. C) $3120$
4. D) $3250$
5. E) $3750$

Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic note, tags and key insights are original to this site.