Question

Twenty-five of King Arthur’s knights are seated at their customary round table. Three of them are chosen, with all choices of three equally likely, and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Answer

Answer: 57

Concepts:combinations complementary counting complementary probability

Difficulty rating: 2410

Small Hint:

Count the complementary selections in which no two chosen knights are adjacent

Big Hint:

Split into selections containing a fixed knight and selections not containing that knight

Solution:

There are $\binom{25}{3}=2300$ selections. Fix one seat. If it is not selected, choosing three nonadjacent seats among the remaining $24$ seats is equivalent to choosing three nonconsecutive positions from a row of $24,$ giving $\binom{22}{3}.$ If the fixed seat is selected, its two neighbors are forbidden, and the other two selected seats must be nonconsecutive among the remaining row of $22,$ giving $\binom{21}{2}.$

Thus the number with no adjacent selected seats is $\begin{aligned} \binom{22}{3}+\binom{21}{2} &=1540+210\\ &=1750. \end{aligned}$ Therefore $P=1-\frac{1750}{2300}=\frac{11}{46},$ and the requested sum is $11+46=57.$

1983 AIME Problem 7

Answer: 57

Difficulty rating: 2410

Small Hint:

Big Hint:

Solution:

Problem 7 in Other Years