AMC 10 · 2024 · #16

Grade 11 probabilitycounting

Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count

combinations-basicfundamental-counting-principleprobability-basicpermutations-basicgcd identify-subproblemssystematic-enumeration ↑ Prerequisites: combinations-basicprobability-basicfraction-arithmetic

📏 Long solution 💡 4 insights

Problem

Twelve tokens — 3 red, 2 white, 1 blue, 6 black — are dealt at random to three players, 4 tokens each. We want the chance that the three coloured groups land in three different hands: one player holds all 3 reds, a second holds both whites, and the third holds the blue. That probability is m/n in lowest terms; report m+n.

Pick an answer.

(A)

387

(B)

388

(C)

389

(D)

390

(E)

391

How to solve

Strategy Identify Subproblems

The event sounds tangled because three things must happen at once, so Tool #7 (Identify Subproblems) splits the count into two independent jobs: first decide which player gets which colour role, then decide which black tokens fill the seats left over. Tool #15 (Organize Information in More Ways) is what makes counting legal at all — treat the 12 tokens as 12 distinct objects, even though same-coloured ones look alike, because only then is every deal equally likely. Tool #9 (Solve an Easier Related Problem) handles the second job: count the black-token fills for one fixed role assignment, which is an easy sequence of choices, and scale up afterwards. Tool #2 (Make a Systematic List) supplies the scale factor by naming all the ways the three roles can be handed out.

1STEP 1

Treat all twelve as different

Distinguishing them makes the counting clean.

12 distinct tokens → hands H₁,H₂,H₃ with |H₁|=|H₂|=|H₃|=4

2STEP 2

Count every possible deal

There are 34650 deals in all.

C(12, 4)C(8, 4)C(4, 4)=495 · 70 · 1=34650

3STEP 3

Pin down what the event requires

Six empty seats match six black tokens exactly.

empty seats=1+2+3=6=number of black tokens

4STEP 4

Hand out the three roles

There are 6 ways to assign the roles.

3!=3 · 2 · 1=6

5STEP 5

Fill the leftover seats with blacks

Splitting blacks one-two-three gives 60 ways.

C(6, 1)C(5, 2)C(3, 3)=6 · 10 · 1=60

6STEP 6

Multiply the two subproblems

That makes 360 good deals.

6 · 60=360 → P=360/34650

7STEP 7

Reduce the fraction and add

The probability is four over 385, so the answer is 389.

360/34650=4/385, m+n=4+385=389 → (C)

Answer

389

The probability 4/385≈ 0.0104 is about one percent, which feels right. Just getting the 3 reds into a single hand already has probability 3·4/12·3/11·2/10=3/55≈ 0.055, and the event demands much more than that, so the answer must be a good deal smaller — it is, by a factor of about five. A stronger check: the deal is small enough to enumerate completely, and running through all 34650 splits of 12 labelled tokens into three hands of 4 finds exactly 360 that satisfy the event, confirming 360/34650=4/385. A structural check also passes: if you refuse to label the players and count unordered groupings instead, the total becomes 34650/6=5775 and the favourable count becomes 360/6=60, and 60/5775=4/385 again — labelling the players cannot change a probability, and it does not. The main trap is forgetting the 3! role assignment, which leaves 60/34650=2/1155 and m+n=1157; that value is not among the choices, which is itself a warning. The second trap is treating same-coloured tokens as identical, which destroys the equally-likely model in the very first step.

💡Key takeaway

When several special groups must land in different piles, first count who gets which job, then count how the leftovers fill the empty seats, and multiply the two counts.

Treat all twelve tokens as different
Count every possible deal
Pin down what the event really requires
Hand out the three roles
Fill the leftover seats with blacks
Multiply the two subproblems
Reduce the fraction and add