Question 1179262: Ten points in the plane are given, with no three collinear. Four
distinct segments joining pairs of these 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 m/n, where m and n
are relatively prime positive integers. Find m + n.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step-by-step:
**1. Total Number of Segments:**
* With 10 points, the number of segments that can be formed is given by the combination formula:
* ¹⁰C₂ = 10! / (2! * 8!) = (10 * 9) / 2 = 45
**2. Total Number of Ways to Choose 4 Segments:**
* The number of ways to choose 4 segments from the 45 available is:
* ⁴⁵C₄ = 45! / (4! * 41!) = (45 * 44 * 43 * 42) / (4 * 3 * 2 * 1) = 148,995
**3. Number of Ways to Choose 4 Segments That DO NOT Form a Triangle:**
To find the probability of forming a triangle, it's easier to find the probability of *not* forming a triangle and subtract that from 1.
For 4 segments not to form a triangle, we need to consider the following cases:
* **Case 1: No intersections.** All four segments are completely separate.
* **Case 2: One intersection.** Two segments intersect at a point, but no triangle is formed.
* **Case 3: Two intersections.** Two sets of intersecting segments, but no triangle is formed.
* **Case 4: All 4 segments form a quadrilateral.**
We will use the complementary counting method.
Consider the cases where we *cannot* form a triangle:
* **Case 1: All 4 segments are disjoint.** This is difficult to calculate directly.
* **Case 2: Choose 4 segments such that no 3 form a triangle.**
Instead, let's look at the complementary case:
* **Case 1: 3 of the 4 segments form a triangle.**
* Choose 3 points out of 10 to form a triangle: ¹⁰C₃ = 120
* Choose 1 remaining segment: 42 segments can be chosen that don't form a triangle with the previous 3.
* So, 120 * 42 = 5040 ways to have 3 segments form a triangle, but we overcount, so we must divide by the number of times we can select the same triangle.
* **Case 2: 4 segments form a complete quadrilateral.**
* Choose 4 points out of 10: ¹⁰C₄ = 210
* Each set of 4 points forms 3 possible quadrilaterals.
* 210 * 3 = 630 ways to have 4 segments form a quadrilateral.
Let's use a different approach.
We need to subtract the cases where we cannot form a triangle.
* **Case 1: 4 disjoint segments.**
* **Case 2: 2 pairs of disjoint segments.**
* **Case 3: A "path" of 4 segments.**
* **Case 4: A "star" with 4 segments.**
Instead, let's find the number of ways to form a triangle.
* Choose 3 points out of 10: ¹⁰C₃ = 120
* Choose 1 more segment from the remaining 42 segments: 42
* Total ways to have a triangle: 120 * 42 = 5040
However, we are overcounting.
We can select the 3 segments of the triangle from the chosen 4 segments in 4 ways.
So we have to divide by 4.
5040/4 = 1260
So there are at least 1260 ways.
Probability = 1260/148995 = 84/9933 = 28/3311.
m=28, n=3311
m+n = 3339
**Final Answer:**
m + n = 28 + 3311 = 3339
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