AMC 10 · 2006 · #18

Grade 7 arithmetic

Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count

permutations-basicsystematic-enumeration identify-subproblems ↑ Prerequisites: permutations-basic

📏 Medium solution 💡 2 insights

Problem

A license plate is a row of 6 characters: 4 digits (each 0–9, repeats allowed) and 2 letters (each A–Z, repeats allowed). The characters can sit in any order, with one rule: the two letters have to be side by side. Count how many different plates you can make.

Pick an answer.

(A)

$10^4\times 26^2$

(B)

$10^3\times 26^3$

(C)

$5\times 10^4\times 26^2$

(D)

$10^2\times 26^4$

(E)

$5\times 10^3\times 26^3$

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The count has three independent choices tangled together: where the letters sit, which letters they are, and which digits fill the rest. Tool #7 (Identify Subproblems) splits the plate into these three separate counts, and the multiplication principle multiplies them. The awkward part is the "letters must be adjacent" rule, so Tool #16 (Change Focus) handles it first: glue the two letters into a single block, which turns a scattered constraint into one solid unit to place. Tool #2 (Make a Systematic List) then counts exactly how many spots that block can occupy. Once the pieces are independent, multiply.

1STEP 1

Glue the two letters into one block

The two letters must touch, so glue them into one block LL — now you are arranging 5 items: the 4 digits plus that block.

6 characters ⟶ 4 digits + 1 LL block = 5 items

2STEP 2

Count where the block can go

The block covers 2 neighboring slots, and the adjacent pairs are (1,2), (2,3), (3,4), (4,5), (5,6) — 5 positions; digits fill the rest.

(1,2), (2,3), (3,4), (4,5), (5,6) → 5 positions

3STEP 3

Count the letters and the digits

The 2 letters give 26 × 26 = 26² ways, and the 4 digit slots give 10 × 10 × 10 × 10 = 10⁴ ways.

letters = 26², digits = 10⁴

4STEP 4

Multiply the three independent counts

Position, letters, and digits are independent, so multiply: 5 × 26² × 10⁴ = 5 × 10⁴ × 26², which is choice (C).

5 × 26² × 10⁴ = 5 × 10⁴ × 26² → (C)

Answer

5 × 10⁴ × 26²

Check the pieces against the answer choices. Every choice contains a 26-power for letters and a 10-power for digits; since there are exactly 2 letters and 4 digits, the value counts must be 26² and 10⁴ — that already rules out B, D, and E, which use the wrong exponents. What remains is whether the leading 5 belongs: choice (A) is 10⁴ × 26² with no factor of 5, which would be the count if the letters were locked into fixed slots. But the plate can slide the letter-block into 5 different positions, so the honest count is 5 times larger. That matches (C) and nothing else.

💡Key takeaway

Tie the two letters into one block so they always stay together, count its 5 possible spots, then multiply by the 10⁴ digit choices and 26² letter choices to get 5 × 10⁴ × 26².

Glue the two letters into one block
Count where the block can go
Count the letters and the digits
Multiply the three independent counts

Glue the two letters into one block

Count where the block can go

Count the letters and the digits

Multiply the three independent counts

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