AMC 10 Step by Step

Arrangements with Restrictions

Permutations with constraints, adjacency, circular arrangements, forbidden positions

25
primary-topic problems (1.9% of all)
19
more as a secondary topic
Where it appears
4
P1-10
3
P11-15
9
P16-20
9
P21-25

What you need to know

  • Block method (glue). To keep kk chosen objects together in a row of nn different objects, tape them into one block. Arrange the n−k+1n-k+1 things in (n−k+1)!(n-k+1)! ways, then multiply by k!k! for the order inside the block. Example: 66 people in a row with Ann, Ben, and Cal together: 4!⋅3!=1444!\cdot3! = 144.
  • Gap method. To keep kk different objects apart (no two side by side), first line up the other mm objects. That leaves m+1m+1 gaps, counting both ends. Put the kk objects into kk different gaps: (m+1k)⋅k!\binom{m+1}{k}\cdot k! ways. Then multiply by the ways to line up the other mm. Example: 44 boys and 33 girls in a row, no two girls together: 4!⋅(53)⋅3!=24⋅10⋅6=14404!\cdot\binom53\cdot3! = 24\cdot10\cdot6 = 1440.
  • Round table. nn different people sit around a table in (n−1)!(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 (n−1)!2\frac{(n-1)!}{2}, for n≥3n\ge3. If the seats are numbered, it is n!n!. Example: 55 people at a round table: 4!=244! = 24 ways; 55 beads on a bracelet: 1212.
  • Identical objects: n!a! b!⋯\frac{n!}{a!\,b!\cdots}, where aa, bb, … count each group of identical objects. Example: LEVEL has two L's and two E's, so 5!2! 2!=30\frac{5!}{2!\,2!} = 30 arrangements.
  • Derangements are arrangements with nothing in its own spot. Call the number for nn objects DnD_n: D2=1D_2 = 1, D3=2D_3 = 2, D4=9D_4 = 9, D5=44D_5 = 44. Example: 44 letters put into 44 addressed envelopes at random are all wrong with probability 94!=924=38\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!n!. Example: 55 people, Ann and Ben not side by side: 120−48=72120-48 = 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≤6n \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) 9696 (B) 144144 (C) 192192 (D) 240240 (E) 288288

Solution. Seat the most limited person first.

Step 1: Ana. She cannot sit at either end, so she takes one of the middle chairs 22, 33, 44, 55: 44 choices.

Step 2: Ben. Every middle chair has a chair on each side, so Ben has 22 choices.

Step 3: everyone else. The other 44 students fill the 44 empty chairs in 4!=244! = 24 ways.

Step 4: multiply. 4⋅2⋅24=1924\cdot2\cdot24 = 192.

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

Pitfalls

  • Forgetting the k!k! for the order inside a block. "Ann, Ben" and "Ben, Ann" are different seatings, so 55 people with those two together give 4!⋅2=484!\cdot2 = 48, not 2424.
  • Using (n−1)!(n-1)! for a circle when the seats are numbered (then it is n!n!), or n!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)

Problems, easiest first

Open any of them to try it here; the worked solution is on the problem's own page.

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