AMC 10 · 2008 · #22

Grade 7 probability
Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count
probability-basicpermutations-basiccasework systematic-enumeration ↑ Prerequisites: probability-basic
📏 Medium solution 💡 3 insights
Problem
Six beads — three red, two white, and one blue — are placed in a line in a random order. Find the probability that no two beads that sit next to each other have the same color.

Pick an answer.

(A)
$\ 1/12$
(B)
$\ 1/10$
(C)
$\ 1/6$
(D)
$\ 1/3$
(E)
$\ 1/2$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

Probability here is just (good arrangements)÷(all arrangements), so the work is careful counting — tool #2 (Make a Systematic List) leads. First count all orderings of the six beads. Then, instead of checking every color pair, tool #7 (Identify Subproblems) trims the job: the lone blue can never sit next to its own color, so 'valid' only means the three reds are kept apart and the two whites are kept apart. Tool #1 (Draw a Diagram) handles the reds — draw six slots and find every way to drop three reds so none touch — and then tool #2 finishes each red pattern by filling the leftover slots with W,W,B while keeping the whites apart. Divide the good count by the total.

1STEP 1

Count all orderings

Same-color beads are interchangeable, so the number of distinct lines is 6!3! 2!\frac{6!}{3!\,2!}, that is 60 — the denominator.

6!/3! 2! 1!=720/12=60
2STEP 2

See which pairs can clash

Blue is alone, so a clash means two reds or two whites touching — keep the reds apart, then keep the whites apart.

3STEP 3

Place three reds so none touch

Number the spots 1 to 6: the reds stay apart in exactly four ways — {1,3,5}\{1,3,5\}, {1,3,6}\{1,3,6\}, {1,4,6}\{1,4,6\}, {2,4,6}\{2,4,6\}.

R{1,3,5}, R{1,3,6}, R{1,4,6}, R{2,4,6}
4STEP 4

Fill the rest with W, W, B

Fill the three free spots with W, W, B and drop the cases where the whites touch: 3+2+2+33+2+2+3 leaves 10 good lines.

3+2+2+3=10
5STEP 5

Divide to get the probability

Out of 60 total lines, 10 are good, so the probability is 1060=16\frac{10}{60}=\frac{1}{6} — choice (C).

10/60=1/6
Answer
1/6
The result 1/6 is a legal probability between 0 and 1 and equals choice (C). It feels right that the chance is small: three of the six beads are red, and forcing all three reds apart in a short line of six is a tight demand, so most random lines fail. A second count confirms the 10: order the three non-red beads first as WBW, WWB, or BWW, which creates four gaps around them; drop the three reds into three different gaps (C(4, 3)=4 ways) to keep reds apart. For WBW the whites are already split, so all 4 work; for WWB and BWW the gap between the two whites must receive a red, leaving C(3, 2)=3 ways each. That is 4+3+3=10, matching the first method.
💡Key takeaway

When beads must avoid same-color neighbors, count the total lines, then pin down the crowded color first and slot the rest into the gaps that stay apart.

  • Count all orderings
  • See which pairs can clash
  • Place three reds so none touch
  • Fill the rest with W, W, B
  • Divide to get the probability

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