King Arthur's knights
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 Ρ be the probability that at least two of the three had been sitting next to each other. If Ρ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?
The answer is 57.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
P(at least two of the three had been sitting next to each other)
= P(all three had been sitting next to each other) + P(two of the three had been sitting next to each other)
= [ n(all three had been sitting next to each other) + n(two of the three had been sitting next to each other) ] / n(any 3 Knights to be chosen out of all 25)
= [ (25 ways, with 25 different middle Knight each time) + { (25 different pairs) × (25 – 2 already chosen pair – 1 to the right of the pair – 1 to the left of the pair, solo Knight) ways } ] / 25C3
= [ 25 + (25 × 21) ] / [ 25! / 3!22! ]
= [ 25 × (1 + 21) × 1 × 2 × 3 ] / [ 25 × 24 × 23 ]
= [ 22 × 6 ] / [ 24 × 23 ]
= 22 / ( 4 × 23 )
= 11 / ( 2 × 23 )
= 11 / 46
Answer = 11 + 46 = 57