Problem:
Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices of three being equally likely - and are sent off to slay a troublesome dragon. Let be the probability that at least two of the three had been sitting next to each other. If is written as a fraction in lowest terms, what is the sum of the numerator and denominator?
Solution:
It is perhaps easier to think in terms of knights, where . We observe first that there are ways of selecting knights if there are no restrictions. But how many of these threesomes include at least two table neighbors?
First, there are ways to pick three neighboring knights. (Consider each knight along with the two knights to his immediate right.) Second, there are ways to pick two neighboring knights (as with three) followed by () ways of picking a third nonneighboring knight. (We must avoid the pair and the two knights on either side.) Thus, there are threesomes that include exactly two neighbors.
Letting be the probability that at least two of the three chosen knights had been neighbors, it follows that
Then and the required sum is .
The problems on this page are the property of the MAA's American Mathematics Competitions