2008 AMC 10B Problem 22

P21-25~5 minPrint

Basic Probability Arrangements with Restrictions Organized listing / direct enumeration Split into exhaustive cases and handle each

Three red beads, two white beads, and one blue bead are placed in line in random order. What is the probability that no two neighboring beads are the same color?

Show solution without answering

Key insight

Place the three red beads first. They cannot touch, so they fit only four ways in six spots. Then check where the white and blue beads can go.

Solution

Beads of the same color look alike, so count color patterns (like RWRBRW). Every pattern is equally likely.

All patterns.

Choose $3$ of the $6$ spots for red: $\tfrac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ ways. Then choose $1$ of the $3$ spots left for blue: $3$ ways. White fills the last two.
Total: $20 \times 3 = 60$ patterns.

Good patterns. The three reds are the hardest to keep apart, so place them first.

3. The reds need three spots from $1$ to $6$ with no two next to each other. The only choices are
$\{1, 3, 5\},\quad \{2, 4, 6\},\quad \{1, 3, 6\},\quad \{1, 4, 6\}.$
4. Reds in $\{1, 3, 5\}$ : the open spots $2, 4, 6$ are not next to each other, so the two whites can never touch. Blue can take any of the $3$ open spots: $3$ patterns.
5. Reds in $\{2, 4, 6\}$ : same reasoning, $3$ patterns.
6. Reds in $\{1, 3, 6\}$ : the open spots are $2, 4, 5$ , and $4, 5$ are neighbors. The whites cannot both go there, so blue must take $4$ or $5$ : $2$ patterns.
7. Reds in $\{1, 4, 6\}$ : the open spots are $2, 3, 5$ , and $2, 3$ are neighbors. Blue must take $2$ or $3$ : $2$ patterns.
8. Good patterns: $3 + 3 + 2 + 2 = 10$ .

The probability is $\dfrac{10}{60} = \dfrac{1}{6}$ .

The answer is $\boxed{\textbf{(C)}\ 1/6}$ .

Why this works

The color with the most beads has the fewest ways to be spread out, so placing it first leaves only a few small cases. Counting color patterns is fine as long as you count the same way on top and bottom of the fraction: every color pattern stands for the same number of orderings of the actual beads ( $3 \times 2 \times 1 \times 2 \times 1 = 12$ ), so the ratio does not change.

Alternative approach

Count real beads instead of patterns, treating all six as different. There are $6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ orderings. Each good color pattern can be filled by the actual beads in $3 \times 2 \times 1$ (reds) $\times\ 2 \times 1$ (whites) $= 12$ ways, giving $10 \times 12 = 120$ good orderings. The probability is $\tfrac{120}{720} = \tfrac16$ , the same answer.

Concise solution

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

Beads of the same color are indistinguishable, so the equally likely outcomes are the color patterns, of which there are
$\frac{6!}{3!\,2!\,1!} = 60.$

The three reds are the tightest constraint: they must occupy three pairwise non-adjacent positions among $1$ through $6$ . The only such triples are $\{1,3,5\}$ , $\{2,4,6\}$ , $\{1,3,6\}$ and $\{1,4,6\}$ .

Reds in $\{1,3,5\}$ or $\{2,4,6\}$ : the remaining positions are also pairwise non-adjacent, so the whites can never touch. The blue may go in any of the $3$ open slots: $3$ patterns each, $6$ total.
Reds in $\{1,3,6\}$ : open slots $2, 4, 5$ , and $4, 5$ are adjacent. The two whites must not both land there, so the blue must be in $4$ or $5$ : $2$ patterns.
Reds in $\{1,4,6\}$ : open slots $2, 3, 5$ ; by the same reasoning the blue is in $2$ or $3$ : $2$ patterns.

Favorable patterns: $6 + 2 + 2 = 10$ , so the probability is $\tfrac{10}{60} = \tfrac{1}{6}$ .

The answer is $\boxed{\textbf{(C)}\ 1/6}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.5)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques