Problem of three numbers

Are possible

C= $\frac{20 \times 19 \times 18}{3 \times 2 \times 1}$ =1140

different choices of 3 integers numbers on [1,20]

Considering all the types of geometric progressions( integers and not integers) the probability would be= $\frac{11}{1140}$

The integers geometric progressions would be: (1; 2; 4), (1; 3; 9),(1; 4; 16); (2; 4; 8), (2; 6; 18), (3; 6; 12), (4; 8; 16) and (5; 10; 20)

And the not integers geometric progressions would be: (4; 6; 9), (8; 12; 18) and (9; 12; 16)

the sum of all is 11

so p= $\frac{11}{1140}$