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Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Number of points and planes.

Ten points in the plane are given with no three collinear. Four distinct segments joining pairs of three points are chosen at random, all such segments being equally likely.The probability that some three of the segments form a triangle whose vertices are among the ten given points is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n.

- is 107
- is 489
- is 840
- cannot be determined from the given information

Number of points

Plane

Probability

But try the problem first...

Answer: is 489.

Source

Suggested Reading

AIME I, 1999, Question 10

Geometry Vol I to IV by Hall and Stevens

First hint

\(10 \choose 3\) sets of 3 points which form triangles,

Second Hint

fourth distinct segment excluding 3 segments of triangles=45-3=42

Final Step

Required probability=\(\frac{{10 \choose 3} \times 42}{45 \choose 4}\)

where \({45 \choose 4}\) is choosing 4 segments from 45 segments

=\(\frac{16}{473}\) then m+n=16+473=489.

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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