1983 AIME Problems/Problem 7
Problem
Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent of 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 the denominator?
Solution
Solution 1
We can apply the complement principle, by finding the probability that none are sitting next to each other, and subtractring from .
Imagine the other (indistinguishable) people are already seated, and fixed into place.
We will place ,
, and
with and without the restriction.
There are places to place
, followed by
places to place
, and
places to place
after
and
. Hence, there are
ways to place
in between these people with restrictions.
Without restrictions, there are places to place
, followed by
places to place
, and
places to place
after
and
. Hence, there are
ways to place
in between these people without restrictions.
Thus, the desired amount is , and the answer is
.
Solution 2
There are configurations for the knights about the table.
There are ways to pick a pair of knights from the trio, and there are
ways to determine which order they are seated. Since these two knights must be attached, we let them be a single entity, so there are
configurations for the entities.
However, this overcounts the instances in which the trio sits together; when all three knights sit together, then two of the pairs from the previous case are counted. However, we only want to count this as one case, so we need to subtract the number of instances in which the trio sits together (as a single entity). There are ways to determine their order, and there are
configurations.
Thus, the answer is , and the answer is
.
See also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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