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2006 AMC 10A Problem 18

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2006 AMC 10A Problem 18AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Counting · difficulty 3 of 5 · about 3 min · position P16-20

A license plate in a certain state consists of 44 digits, not necessarily distinct, and 22 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?

  1. A)

    104×26210^4\times 26^2

  2. B)

    103×26310^3\times 26^3

  3. C)

    5×104×2625\times 10^4\times 26^2

  4. D)

    102×26410^2\times 26^4

  5. E)

    5×103×2635\times 10^3\times 26^3

Answer

C

Key insight

Glue the two adjacent letters into one block: it can start in any of 5 positions among the 6 slots, times 10^4 digit strings times 26^2 letter pairs.

Solution

Because the two letters are adjacent, treat them as one two-character block. A plate is then a row of five items: the block and four digit positions.

Choose where the block sits: its left end can be slot 1,2,3,41, 2, 3, 4, or 55 of the six slots, so 55 positions.

Fill the block: 2626 choices for the first letter and 2626 for the second (repeats allowed and the order is part of the plate), so 26226^2 ordered pairs.

Fill the four digit slots from left to right: 1010 choices each, 10410^4 strings.

By the multiplication principle the total is 5×104×2625 \times 10^4 \times 26^2.

The answer is (C) 5×104×262\boxed{\textbf{(C)}\ 5\times 10^4\times 26^2}.

Why this works

An adjacency restriction is handled by bundling the adjacent items into a block, which converts the restriction into an ordinary placement count. Since digits and letters are read as strings (order already built in), no factorials are needed: just count block positions, then fill every slot independently.

Alternative approach

Answer-choice check: the plate certainly involves 10410^4 digit choices and 26226^2 letter choices, so the count is 104⋅26210^4 \cdot 26^2 times the number of layouts. Choice (A) would mean only one layout, which is false; the only other choice of that shape is (C).

The trap

Multiplying by an extra 2 for the letters' order (26^2 already distinguishes AB from BA) or by 6 instead of 5 for the block's position.

Common mistakes

  • Multiplying by an extra 22 for the letters' order (26226^2 already distinguishes AB from BA) or by 66 instead of 55 for the block's position.
  • Multiplying by 4!4! to "arrange" the digits, which double counts because 10410^4 already lists every digit string in order.

Techniques

Set up the equation/formula and compute; no special trick needed