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 Geometric Generalization (Posted on 2013-02-28)
Three positive integers are chosen at random without replacement from 1,2,....,64. What is the probability that the numbers chosen are in geometric sequence?

Order of choice doesn't matter. For example, 4-1-2 would qualify as numbers in geometric sequence.

Bonus Question:
Generalise this result (in terms of n) covering the situation where three positive integers are chosen at random without replacement from 1,2,.....,n2.

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution and most of bonus | Comment 2 of 6 |
First note that there are C(64,3)=41664 unordered ways of choosing 3 numbers from the 64.  The real question is how many of these triplets are in geometric sequence.

If the common ratio of terms r=2 there are 64/2^2 = 16 triplets
if r=3 there are 64/3^2=7.111 so 7 triplets
if r=4 there are 4 triplets
if r=5 there are 2
if r=6, 7 or 8 there is only 1 each.
The total is 32 unordered triples.
***EDIT: THIS IS WRONG.  I DID NOT CONSIDER NON-INTEGER RATIOS***
32/41664 = 1/1302 ≈ .000768

Bonus:  You can basically count up the same way there are C(nē,3) = nē(nē-1)(nē-2)/3 unordered ways of choosing 3 numbers.
Then count the unordered triples as before
Sum [nē/rē] for all r from 2 to n
Then divide the two.
I'll work on the closed form for this.
***WHICH MEANS THIS IS ALSO WRONG***

Edited on February 28, 2013, 10:57 am
 Posted by Jer on 2013-02-28 10:54:38

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