AMC 10 · 2021 · #18

Grade 11 probability
Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count
probability-basiccombinations-basicfactorialcombinatorial-identity identify-subproblemseasier-related-problem ↑ Prerequisites: probability-basiccombinations-basic
📏 Long solution 💡 3 insights
Problem
Twenty balls are thrown one at a time, each landing in one of five bins with equal chance and independently. Let the first probability be that the final counts are one bin with 3, one with 5, and the other three with 4 each. Let the second be that all five bins finish with exactly 4. Find the first divided by the second.

Pick an answer.

(A)
1
(B)
4
(C)
8
(D)
12
(E)
16
How to solve
Strategy Change Focus / Count the Complement

Tool #16 (Change Focus) is the whole game here: the question asks for a ratio of two probabilities, not for either probability, and both probabilities are counts out of the same 5²⁰ equally likely outcomes. Shifting attention from probabilities to raw outcome counts makes the denominator vanish before any work starts. Tool #7 (Identify Subproblems) then splits each count into two independent decisions — which bins get the unusual sizes, and which balls go where — so each count is a short product instead of a case bash. Tool #9 (Solve an Easier Related Problem) keeps the numbers small: the counts involve 20!, which never has to be evaluated, because writing the ratio as a single fraction lets the huge factorials cancel and leaves one line of arithmetic.

1STEP 1

Same denominator, so compare counts

Same denominator, so compare only the counts.

p = N_p/5²⁰, q = N_q/5²⁰ ⟹ p/q = N_p/N_q
2STEP 2

Count the all-fours outcomes

Count the all-fours outcomes.

N_q = C(20, 4)C(16, 4)C(12, 4)C(8, 4)C(4, 4) = 20!/4! 4! 4! 4! 4! = 20!/(4!)⁵
3STEP 3

Count the 3-5-4-4-4 outcomes

You must also choose which bins differ.

N_p = 5 · 4 · 20!/3! 5! 4! 4! 4! = 20 · 20!/3! 5! (4!)³
4STEP 4

Cancel before multiplying

Cancel before multiplying.

N_p/N_q = 20 · 20!/3! 5! (4!)³ · (4!)⁵/20! = 20 · (4! · 4!)/(3! · 5!) = 20 · 4!/3! · 4!/5!
5STEP 5

Finish the arithmetic

The arithmetic gives 16.

20 · 4!/3! · 4!/5! = 20 · 4 · 1/5 = 16 ⟹ (E)
Answer
16
The direction is right and the size is right. Compare a single fixed labeling first: with the small bin and the large bin pinned down, the lopsided pattern is actually the rarer one, since (4! · 4!)/(3! · 5!) = 4/5 of the all-fours count. But there are 20 ways to pin those two bins down against only 1 way to be perfectly even, and 20 · 4/5 = 16 beats 1 comfortably — which also rules out (A) 1, the answer one would get by wrongly assuming every pattern is equally likely. A small case confirms the method: throw 4 balls into 2 bins, and the same reasoning predicts 2 · (2! · 2!)/(1! · 3!) = 4/3; counting by hand agrees, since the (1,3) pattern has 2 · 4!/1! 3! = 8 outcomes against 4!/2! 2! = 6 for (2,2), and 8/6 = 4/3.
💡Key takeaway

Both events are counted out of the same 5²⁰ equally likely outcomes, so the ratio of probabilities is just a ratio of counts — write it as one fraction and the giant factorials cancel each other away.

  • Same denominator, so compare counts
  • Count the all-fours outcomes
  • Count the 3-5-4-4-4 outcomes
  • Cancel before multiplying
  • Finish the arithmetic

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