AMC 10 · 2025 · #13

Grade 7 probabilitycounting

Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements

probability-basiccombinations-basicconsecutive-integer-count extremal-constructionsystematic-enumeration ↑ Prerequisites: combinations-basic

📏 Medium solution 💡 3 insights

Problem

The set is C = {1, 2, 3, ..., 13}. First find N, the largest size of a subset of C that contains no five consecutive integers (no block like k, k+1, k+2, k+3, k+4 all present). Then N numbers are drawn at random from C without replacement, and we want the probability that the drawn numbers still contain no five consecutive integers.

Pick an answer.

(A)

$\frac{3}{130}$

(B)

$\frac{3}{143}$

(C)

$\frac{5}{143}$

(D)

$\frac{1}{26}$

(E)

$\frac{5}{78}$

How to solve

Strategy Change Focus / Count the Complement

Keeping N numbers out of 13 is the same as leaving a few out, and the whole problem gets simpler when I focus on the numbers left out instead of the numbers kept. To make the kept set as large as possible I make the left-out set as small as possible, so I first ask how few numbers I can remove and still break every run of five. Once that shows only two numbers need to go, both the sample space and the favorable outcomes are naturally counted by the pair of numbers left out, turning a scary subset problem into counting a handful of pairs.

1STEP 1

Keep the most by removing the fewest

Remove one number and the two leftover runs can't both stay under five, so at least two numbers must go.

N = 13 - (numbers removed); k-1 ≤ 4 and 13-k ≤ 4 → k ≤ 5 and k ≥ 9

2STEP 2

Two removals split the line into three blocks

Removing two numbers cuts the line into three blocks; each under five means a at most 5, b−a at most 5, b at least 9.

a-1 ≤ 4, b-a-1 ≤ 4, 13-b ≤ 4 → a ≤ 5, b-a ≤ 5, b ≥ 9

3STEP 3

List the removal pairs that work

Only (4,9), (5,9), (5,10) satisfy all three, so two removals suffice and N = 11.

(4,9), (5,9), (5,10) → N = 13 - 2 = 11

4STEP 4

Count all equally likely draws

Choosing 11 to keep equals choosing 2 to leave out, so there are C(13,2) = 78 equally likely outcomes.

C(13, 11)=C(13, 2)=(13 · 12)/2=78

5STEP 5

Divide favorable by total

The three winning pairs are exactly the favorable leave-outs, so the probability is $\frac{3}{78}$ = $\frac{1}{26}$ , choice (D).

P=3/78=1/26

Answer

1/26

The probability

\frac{1}{26}

is about 0.038, a small chance, which fits: only three of the 78 ways to leave out two numbers manage to cut every possible run of five, so success should be rare. The value sits strictly between 0 and 1, as any probability must. A quick recheck confirms each winning subset has longest run 4: leaving out {4,9} gives block lengths 3,4,4; leaving out {5,9} gives 4,3,4; leaving out {5,10} gives 4,4,3, all at most 4.

💡Key takeaway

Choosing which few to leave out is often easier than choosing which many to keep, and a probability is just the winning outcomes divided by all the equally likely outcomes.

Keep the most by removing the fewest
Two removals split the line into three blocks
List the removal pairs that work
Count all equally likely draws
Divide favorable by total

Keep the most by removing the fewest

Two removals split the line into three blocks

List the removal pairs that work

Count all equally likely draws

Divide favorable by total

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