Counting Without Missing or Repeating

01 · The one rule behind every method

A correct count gives every valid outcome one name, exactly once. Most counting errors come from missing an outcome or naming the same outcome more than once.

02 · Build the formulas instead of memorizing them

The product rule

Suppose a meal has 3 choices of main dish, 4 choices of a pair of vegetables, and 2 choices of dessert. The stages are independent:

This pattern is adapted from the choice structure in 2001 AMC 8, Problem 14.

Permutations

For 4 different objects, the first position has 4 choices, then 3, then 2, then 1:

Combinations

Choosing 2 from 4 first produces 4 × 3 ordered choices. Every pair appears twice, once in each order, so divide by 2!:

03 · The order test

Ask one question: If I swap two chosen objects, do I get a new outcome?

04 · Count all, then subtract the bad

Four different marbles are placed in a row. Two special marbles may not be next to each other.

1. Count all: 4! = 24.
2. Glue the forbidden pair: the block plus two other marbles gives 3! orders.
3. Flip inside the block: 2 orders.
4. Subtract: 24 − 2·3! = 12.

This is an adapted version of the central idea in 2020 AMC 8, Problem 10. The same “block” idea appears in harder bookshelf problems such as 2018, Problem 16.

05 · Repeated objects and gap counting

If a word contains repeated letters, swapping identical copies changes nothing. For the letters in LEVEL, there are 5 letters with two L's and two E's:

For “no two special letters together,” arrange the ordinary objects first and use the gaps around them.

Four objects create five gaps. Placing one identical special object in every gap avoids adjacency. This gap structure is the key to 2022 AMC 8, Problem 14.

06 · Turn paths into strings

A shortest path that moves 3 blocks east and 2 blocks north is a string containing EEE and NN. Choose where the two N moves go:

If one intersection is forbidden, subtract paths through it. To pass through (1,1) on the way from (0,0) to (3,2):

This method tracks the structure of 2014 AMC 8, Problem 11. Path counting also appears in 1986 #9, 2013 #21, 2017 #15, and 2020 #21.

07 · When a recurrence is better than a formula

Let f(n) count ways to climb n stairs using jumps of 1, 2, or 3. Every valid route ends with exactly one of those jumps:

This is the engine behind 2010 AMC 8, Problem 25. A similar state-building idea appears in 2024 #13 and 2026 #20.

08 · A map of genuine AMC 8 patterns

The private archive was searched across 41 contests. These references show how the same small toolkit reappears in different disguises.

The historical references identify source patterns. Problems and worked examples on this page are rewritten or newly created rather than copied verbatim from MAA tests.