Can you do it without PIE ?
Twenty - five of King Arthur's knights are seated at their customary roundtable. 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
The answer is 57.
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1 solution
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