AMC 10 · 2005 · #21

Grade 7 probability

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

combinations-basicprobability-basic identify-subproblemsratio-proportion ↑ Prerequisites: combinations-basicprobability-basic

📏 Medium solution 💡 2 insights

Problem

A hat holds forty slips: the numbers 1 through 10, each printed on exactly four slips. Draw four slips at random without replacement. Let p be the chance all four show the same number, and q the chance two show one number a and the other two show a different number b. Find q/p.

Pick an answer.

(A)

162

(B)

180

(C)

324

(D)

360

(E)

720

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Both p and q are (favorable draws) divided by the same total number of four-slip draws, C(40, 4). Tool #16 (Change Focus) turns the whole problem on this fact: in the ratio q/p that shared total cancels, so q/p equals just the count of two-pair draws divided by the count of four-of-a-kind draws — the giant C(40, 4) never has to be computed. That leaves two clean counting subproblems (Tool #7), each handled by Tool #2 (Make a Systematic List): count the four-of-a-kind draws, then count the two-and-two draws, and divide.

1STEP 1

Both chances share one total

Every four-slip draw is equally likely, so p and q both sit over C(40, 4) — that shared total cancels in q/p, leaving count over count.

q/p = ( N_q/C(40, 4) )/( N_p/C(40, 4) ) = N_q/N_p

2STEP 2

Count four-of-a-kind draws

All four alike means one number's whole set of four, so only the number is free to choose: N_p = 10.

N_p = 10 × C(4, 4) = 10 × 1 = 10

3STEP 3

Count two-and-two draws

Pick the unordered pair a, b in 45 ways, then two slips from each of their fours in 6 × 6 ways: N_q = 1620.

N_q = C(10, 2)C(4, 2)C(4, 2) = 45 × 6 × 6 = 1620

4STEP 4

Divide the two counts

Substitute into q/p = N_q/N_p: 1620 divided by 10 gives 162, a clean whole number — choice (A).

q/p = N_q/N_p = 1620/10 = 162 → (A)

Answer

162

The value 162 is a whole number, which fits: because p and q share the denominator C(40, 4), the ratio had to reduce to N_q/N_p with N_p=10 dividing evenly into N_q=1620. It is also reasonable that q is far larger than p — a two-and-two split can happen in many more ways (1620) than a rare perfect match (10), so q/p ≫ 1. The trap answers come from miscounting: forgetting that {a,b} is unordered doubles 1620 to 3240 and gives 324 (C); using 2C(10, 2) style slips-order overcounts toward 360 (D) or 720 (E). Only (A) matches the careful count.

💡Key takeaway

Since p and q are both counts over the same total, q/p is just one count over the other: 1620/10=162.

Both chances share one total
Count four-of-a-kind draws
Count two-and-two draws
Divide the two counts

Both chances share one total

Count four-of-a-kind draws

Count two-and-two draws

Divide the two counts

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