2006 AMC 10A Problem 18

P16-20~3 minPrint

Basic Counting Arrangements with Restrictions Set up the equation/formula and compute; no special trick needed

A 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?

Show solution without answering

Key insight

Glue the two letters into one block. It can start in 5 of the 6 spots; fill it with 26 × 26 letter pairs and the rest with 10^4 digits.

Solution

The letters must be side by side, so treat the pair of letters as one block that takes up two neighboring spots.

1. Where the block goes. The plate has $6$ spots. The block can take spots $1$ – $2$ , $2$ – $3$ , $3$ – $4$ , $4$ – $5$ or $5$ – $6$ . That is $5$ ways.
2. The letters in the block. $26$ choices for the first letter and $26$ for the second (repeats allowed), so $26^2$ ways. This already counts "AB" and "BA" as different.
3. The digits. The other $4$ spots each get one of $10$ digits: $10^4$ ways.
4. Multiply. Each choice in one step can go with each choice in the others:
$5 \times 10^4 \times 26^2.$

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

Why this works

When some items must stay next to each other, glue them into one block. Then the rule disappears: you only count where the block can go and how to fill every spot. Because the plate is read left to right, the order is already built into the filling, so no extra factor is needed.

Alternative approach

Look at the choices. Any plate uses $4$ digits and $2$ letters, so the count must be $10^4 \times 26^2$ times the number of ways to arrange where the letters go. Choice (A) would mean only one arrangement, but "letters first" and "letters last" are already two. The only other choice of the form $10^4 \times 26^2 \times (\text{something})$ is (C).

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $1, 2, 3, 4$ , or $5$ of the six slots, so $5$ positions.

Fill the block: $26$ choices for the first letter and $26$ for the second (repeats allowed and the order is part of the plate), so $26^2$ ordered pairs.

Fill the four digit slots from left to right: $10$ choices each, $10^4$ strings.

By the multiplication principle the total is $5 \times 10^4 \times 26^2$ .

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.5)
Introduction to Counting & Probability, Chapter 3: Correcting for Overcounting (§3.2, §3.4)
Introduction to Counting & Probability, Chapter 4: Committees and Combinations