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### Probability of Forming a Triangle

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Date: 08/09/99 at 09:38:24
From: Sharon Lawson
Subject: Probability and Triangles

Hi,

I wonder if anybody can help me solve the following problem:

Three numbers are chosen from the first n natural numbers. What is the
probability that they can be sides of a triangle? The numbers are
picked "without replacement," so we cannot pick the same number twice.

I know that for this to be possible, the sum of any two sides must be
greater than the third side, and obviously the sum of the three
numbers cannot exceed 3*n - 3, but I don't know how to do it from
there.

Thanks in advance for any help.
Sharon
```

```
Date: 08/09/99 at 18:47:36
From: Doctor Anthony
Subject: Re: Probability and Triangles

If the numbers range from 1 to n we can find a pattern for numbers
that will form a triangle. In the pattern that I derive below, the
case where the two shorter sides sum to give the longest side is
included as a collapsed triangle. I give the two shorter sides first,
then the longest side.

1+2     - 3     2+3     - 4,5         3+4     - 5,6,7
1+3     - 4     2+4     - 5,6         3+5     - 6,7,8
1+4     - 5     2+5     - 6,7         3+6     - 7,8,9
:        :      :         :           :          :
1+(n-1) - n     2+(n-2) - (n-1),n     3+(n-3) - (n-2),(n-1),n
------------    2+(n-1) - n           3+(n-2) - (n-1),n
Total = n-2     ------------------    3+(n-1) - n
Total = 2n-7        ------------------
Total  = 3n-15

Continuing in this way we get the following series:

(n-2), (2n-7), (3n-15), (4n-26), (5n-40), (6n-57), ...

The numbers in the brackets 2, 7, 15, 26, 40, 57, ... are given by:

r(3r+1)/2   starting at r = 1

The end of the sequence is more complicated and we have different
results depending on whether n is odd or even.

The top limit for r must be restricted to (n-1)/2 when n is odd and
(n-2)/2 when n is even. Thereafter we have the sum of the triangular
numbers up to the (n-3)/2 such number (for n odd) and the (n-2)/2 such
number (for n even).

The triangular numbers are 1, 3, 6, 10, 15, 21, ... and the nth number
is given by:

n(n+1)/2

The sum of n terms of this triangular number series is given by:

(1/2)SUM[n^2 + n]

= (1/2)[n(n+1)(2n+1)/6 + n(n+1)/2]

= (1/4)n(n+1)[(2n+1)/3 + 1]

= (1/12)n(n+1)[2n+1+3]

= (1/12)n(n+1)[2n+4]

= (1/6)n(n+1)(n+2)

So the formula for the number of triangles will be, for n odd:

Let k = (n-1)/2

k
SUM[r.n - r(3r+1)/2] + (1/6)(k-1)k(k+1)
r=1

k
SUM[r.n - r(3r+1)/2] + (1/6)k(k^2-1)
r=1

And the formula for the number of triangles with n even:

Let k = (n-2)/2

k
SUM[r.n - r(3r+1)/2] + (1/6)k(k+1)(k+2)
r=1

Now you could choose 3 numbers from n numbers in C(n,3) ways, so the
probability of forming a triangle, for n odd, is given by:

k
SUM[r.n - r(3r+1)/2] + (1/6)k(k^2-1)
r=1
--------------------------------------   where k = (n-1)/2
C(n,3)

and for n even the probability is:

k
SUM[r.n - r(3r+1)/2] + (1/6)k(k+1)(k+2)
r=1
-----------------------------------------   where k = (n-2)/2
C(n,3)

To avoid tedious algebra we shall work out this probability in the
case where n = 20.

k = (n-2)/2 = 9

The top line is:

9
SUM[20r - r(3r+1)/2] + (1/6)k(k+1)(k+2)
r=1

= 20.SUM[r) - (1/2)SUM[3r^2] - (1/2)SUM[r] +(1/6)(9)(10)(11)

for r = 1 to 9

= 20(9)(10)/2 - (1/2)(3)(9)(10)(19)/6 - (1/2)(9)(10)/2 + 165

= 900 - 427.5 - 22.5 + 165

=  615

and C(20,3) = 1140

and so the required probability is  615/1140 = 41/76

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

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