2008 AMC 10B Problem 22

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

All patterns.

1. 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.
2. 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 $\boxed{\textbf{(C)}\ 1/6}$.

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.

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.

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