2008 AMC 10B Problem 22

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

When one color is the majority, its non-adjacent placements are few, and fixing them turns the rest into a small check. Counting patterns rather than labeled beads is legitimate as long as numerator and denominator use the same convention: each color pattern corresponds to the same number ($3!\,2!\,1! = 12$) of labeled orderings.

Think of the string as three reds with the other three beads in the gaps around them. Three non-touching reds in six slots leave exactly one gap that holds two beads (or none, in the two "alternating" cases). The count $4$ patterns of red positions and the blue-in-the-double-gap condition reproduce $10$ quickly.

• Using 6! = 720 as the denominator while counting color patterns in the numerator, instead of 60 color patterns for both.
• Forgetting that the whites must also be separated, which counts $4 \cdot 3 = 12$ favorable patterns and gives $\tfrac15$.