2008 AMC 10B Problem 22

Question

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

\dfrac{1}{12}

\dfrac{1}{10}

\dfrac{1}{6}

\dfrac{1}{3}

\dfrac{1}{2}

Answer: C

Concepts:basic probability multiset permutations arrangements with restrictions

Difficulty rating: 1680

Small hint:

There are

\dfrac{6!}{3!\,2!}=60

distinguishable orderings

Big hint:

First place the three red beads so no two are adjacent, then fit the white and blue beads

Solution:

There are

\tfrac{6!}{3!\,2!}=60

distinguishable orderings. The three reds must occupy non-adjacent positions, and the possible red placements are

\{1,3,5\},\{2,4,6\},\{1,3,6\},

and

\{1,4,6\}.

For

\{1,3,5\}

and

\{2,4,6\},

the remaining seats are mutually non-adjacent, so the blue bead can go in any of the

3,

giving

3+3=6.

For

\{1,3,6\}

and

\{1,4,6\},

two remaining seats are adjacent, so the blue must separate the whites, giving

2+2=4.

That is

10

valid orderings, so the probability is

\tfrac{10}{60}=\tfrac16.

Thus, the correct answer is C.