AMC 10 · 2008 · #22
Grade 7 probabilityThree 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?
Pick an answer.
AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The beads are $3$ red ($R$), $2$ white ($W$), and $1$ blue ($B$).; They are arranged in a straight line, and every ordering is equally likely.; Beads of the same color are identical to each other.
Unknowns: The probability that no two neighboring beads share a color.
Understand
Restated: 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.
Givens: The beads are $3$ red ($R$), $2$ white ($W$), and $1$ blue ($B$).; They are arranged in a straight line, and every ordering is equally likely.; Beads of the same color are identical to each other.
Plan
Primary tool: #2 Make a Systematic List
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
Probability here is just (good arrangements)$\div$(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.
Execute — Answer: C
7.SP.C.8 Step 1 Count all orderings
- The six beads come in identical groups, so the number of distinct lines is $6!$ divided by $3!$ (for the identical reds) and $2!$ (for the identical whites).
- That gives $\frac{720}{6\times 2}=60$ equally likely arrangements to serve as the denominator.
💡 Swapping two identical beads does not make a new line, so divide out the ways each color group can be shuffled among itself.
7.SP.C.7 Step 2 See which pairs can clash
- Same-color neighbors can only happen with two reds together or two whites together, because there is just one blue bead.
- So a line is valid exactly when no two $R$'s are adjacent and no two $W$'s are adjacent.
- Split the counting into two easier jobs: first place the reds apart, then place the whites apart.
💡 The single blue never fights anyone, so the only rules to enforce are 'reds apart' and 'whites apart'.
7.SP.C.8 Step 3 Place three reds so none touch
- Label the six spots $1$ to $6$ and choose three of them for the reds so that no two chosen spots are next to each other.
- Listing them systematically gives exactly four patterns: $R$ at $\{1,3,5\}$, $\{1,3,6\}$, $\{1,4,6\}$, and $\{2,4,6\}$.
- (Think of it as sliding three reds into the gaps around the other three beads: $\binom{4}{3}=4$.)
💡 Reds only stay apart if they land in every-other-ish slots, and there are just four such spacings.
7.SP.C.8 Step 4 Fill the rest with W, W, B
- In each red pattern, the three empty spots get $W,W,B$.
- There are $3$ ways to place them ($B$ picks one of the three spots, the whites take the other two), but a placement is bad only when the two whites land on adjacent spots.
- Patterns $R\{1,3,5\}$ and $R\{2,4,6\}$ leave empty spots that are all separated, so all $3$ placements work.
- Patterns $R\{1,3,6\}$ (empties $2,4,5$) and $R\{1,4,6\}$ (empties $2,3,5$) each have one adjacent pair of empties, so $1$ of the $3$ placements puts the whites together and is thrown out, leaving $2$.
- The good total is $3+2+2+3=10$.
💡 Once the reds are set, the only remaining danger is the two whites landing side by side, which is easy to subtract off.
7.SP.C.7 Step 5 Divide to get the probability
- There are $10$ good arrangements out of $60$ total, so the probability is $\frac{10}{60}=\frac{1}{6}$.
- This matches choice $(C)$.
💡 Probability of a fair setup is just the count of wins over the count of everything.
7.SP.C.8 The six beads come in identical groups, so the number of distinct lines is $6!$ 7.SP.C.7 Same-color neighbors can only happen with two reds together or two whites togeth 7.SP.C.8 Label the six spots $1$ to $6$ and choose three of them for the reds so that no 7.SP.C.8 In each red pattern, the three empty spots get $W,W,B$. There are $3$ ways to pl 7.SP.C.7 There are $10$ good arrangements out of $60$ total, so the probability is $\frac Review
Reasonableness: The result $\frac{1}{6}$ 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 ($\binom{4}{3}=4$ 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 $\binom{3}{2}=3$ ways each. That is $4+3+3=10$, matching the first method.
Alternative: Instead of placing reds first, place the non-red beads first and insert reds into the gaps (tool #16, change focus). The three beads $W,W,B$ have $3$ orderings, each making $4$ gaps; choosing $3$ gaps for the reds keeps reds apart, and requiring a red in the gap between two adjacent whites keeps whites apart. Summing the valid insertions gives the same $10$ good lines out of $60$, hence $\frac{1}{6}$.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Counting the $60$ total orderings and systematically counting the $10$ valid ones by placing reds apart and then the whites.)7.SP.C.7Develop a probability model and use it to find probabilities of events (Treating all $60$ orderings as equally likely and computing the probability as favorable over total, $\frac{10}{60}=\frac{1}{6}$.)
⭐ 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.
⭐ 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.
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