Question

Let $C = \{1, 2, 3, \ldots, 13\}.$ Let $N$ be the greatest integer such that there exists a subset of $C$ with $N$ elements that does not contain five consecutive integers. Suppose $N$ integers are chosen at random from $C$ without replacement. What is the probability that the chosen elements do not include five consecutive integers?

$\dfrac{3}{130}$

$\dfrac{3}{143}$

$\dfrac{5}{143}$

$\dfrac{1}{26}$

$\dfrac{5}{78}$

Answer: D

Concepts:combinations basic probability extremal argument

Difficulty rating: 1660

Solution:

To avoid five consecutive integers, it suffices to remove two elements (for example $5$ and $10$ ), and no single removal can hit both disjoint blocks $\{1,\ldots,5\}$ and $\{9,\ldots,13\}.$ Thus $N = 11.$

Choosing $11$ of $13$ elements is the same as removing $2,$ which can be done in $\binom{13}{2} = 78$ ways. The chosen set avoids five consecutive integers exactly when the two removed elements together intersect every window $\{t, t+1, t+2, t+3, t+4\}$ for $t = 1, \ldots, 9.$

This forces one removed element in $\{1,\ldots,5\},$ the other in $\{9,\ldots,13\},$ and the two within $5$ of each other. The valid removals are $\{4,9\},$ $\{5,9\},$ and $\{5,10\},$ giving $3$ of them.

The probability is $\dfrac{3}{78} = \dfrac{1}{26}.$

Thus, the correct answer is D.

2025 AMC 12A Problem 13

Answer: D

Difficulty rating: 1660

Solution:

Problem 13 in Other Years