AMC 10 · 2006 · #18
Grade 7 arithmeticA license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. These six characters may appear in any order, except that the two letters must appear next to each other. How many distinct license plates are possible?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: $4$ digit characters, each independently $0$–$9$; $2$ letter characters, each independently $A$–$Z$; The $6$ characters fill $6$ slots in a row; The two letters must occupy two adjacent slots; Answer choices: (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) $5\times 10^3\times 26^3$
Unknowns: The total number of distinct license plates
Understand
Restated: 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.
Givens: $4$ digit characters, each independently $0$–$9$; $2$ letter characters, each independently $A$–$Z$; The $6$ characters fill $6$ slots in a row; The two letters must occupy two adjacent slots; Answer choices: (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) $5\times 10^3\times 26^3$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #16 Change Focus / Count the Complement, #2 Make a Systematic List
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.
Execute — Answer: C
7.SP.C.8 Step 1 Glue the two letters into one block
- The two letters must be adjacent, so treat them as a single glued block, $\boxed{LL}$.
- Instead of $6$ loose characters, you are now arranging $5$ items in a row: the $4$ digits and this one letter-block.
- Handling the adjacency rule up front means every later choice is free — nothing else has a restriction.
💡 Bundling the two letters into one piece makes "they must touch" automatic instead of a rule you keep checking.
7.SP.C.8 Step 2 Count where the block can go
- The letter-block covers $2$ neighboring slots out of the $6$.
- List the adjacent pairs it can occupy: slots $(1,2)$, $(2,3)$, $(3,4)$, $(4,5)$, $(5,6)$.
- That is $5$ possible positions for the block; the $4$ digits then fill whatever slots are left over.
💡 A block two-wide has one fewer place to start than the strip is long, so $6$ slots give $5$ starting spots.
6.EE.A.1 Step 3 Count the letters and the digits
- Now fill in values.
- The block holds $2$ letters; each is one of $26$, chosen independently, so $26 \times 26 = 26^2$ ways.
- The $4$ remaining slots hold digits; each is one of $10$, chosen independently, so $10 \times 10 \times 10 \times 10 = 10^4$ ways.
- Each factor is a repeated multiplication, which is exactly what an exponent records.
💡 Independent choices multiply, and multiplying the same number of options over and over is what a power means.
7.SP.C.8 Step 4 Multiply the three independent counts
- Position of the block, the letters inside it, and the digits around it are three choices that do not affect one another, so multiply them: $5 \times 26^2 \times 10^4$.
- Reordering the factors gives $5 \times 10^4 \times 26^2$, which is choice $\textbf{(C)}$.
💡 When choices are made independently, the total number of combinations is just their counts multiplied together.
7.SP.C.8 The two letters must be adjacent, so treat them as a single glued block, $\boxed 7.SP.C.8 The letter-block covers $2$ neighboring slots out of the $6$. List the adjacent 6.EE.A.1 Now fill in values. The block holds $2$ letters; each is one of $26$, chosen ind 7.SP.C.8 Position of the block, the letters inside it, and the digits around it are three Review
Reasonableness: 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 $26^2$ and $10^4$ — that already rules out B, D, and E, which use the wrong exponents. What remains is whether the leading $5$ belongs: choice (A) is $10^4\times 26^2$ 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.
Alternative: Skip the gluing and count positions directly. Choose which $2$ of the $6$ slots are the letters, but only adjacent pairs are allowed — there are $5$ such pairs. Fill those $2$ slots with letters ($26^2$ ways) and the other $4$ slots with digits ($10^4$ ways). Multiplying gives $5\times 26^2\times 10^4$, the same result. As a cross-check, if adjacency were dropped entirely, the letters could take any $2$ of $6$ slots — $\binom{6}{2}=15$ pairs — so the adjacency rule cuts $15$ down to $5$, exactly a factor of $3$ fewer plates, which fits the picture of most slot-pairs being non-adjacent.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Organizing the plate into independent sub-counts and applying the multiplication principle: $5$ block positions, times $26^2$ letters, times $10^4$ digits.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Recording the repeated independent choices as powers — $26 \times 26 = 26^2$ for the two letters and $10\times10\times10\times10 = 10^4$ for the four digits.)
⭐ Tie the two letters into one block so they always stay together, count its $5$ possible spots, then multiply by the $10^4$ digit choices and $26^2$ letter choices to get $5\times 10^4\times 26^2$.
⭐ Tie the two letters into one block so they always stay together, count its $5$ possible spots, then multiply by the $10^4$ digit choices and $26^2$ letter choices to get $5\times 10^4\times 26^2$.
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