Sets

Let's consider all possible three-element subsets: ${25\choose 3}$ We will now eliminate all the triplets with a common factor. We'll only consider the triplets with a prime common factor since all triplets with a composite common factor are also triplets with a prime common factor. Let's consider the case in which the common factor is 2, we have to find the number of triplets made only by even numbers. Since there are only 12 even numbers between 1 and 25, the number of even triplets is ${12\choose 3}$ . Now let's find the number of triplets with the common factor three. Since there are 8 numbers divisible by 3 between 1 and 25, the number of those triplets is ${8\choose 3}$ . There's only one problem, some triplets with the common factor 3 are already included in the even triplets, these triplets would have a common factor of 6, so by removing them from ${8\choose 3}$ , we get ${8\choose 3}$ - ${4\choose 3}$ . Great, now, applying the same reasoning to triplets with common factor 5 we get ${5\choose 3}$ , and for triplets, with common factor 7, we get ${3\choose 3}$ . There aren't any triplets with common factor 11, since there are only 2 multiples of 11 between 1 and 25, same goes for bigger primes.

Let's now combine all of our previous results:

${25\choose 3}- {12\choose 3}-({8\choose 3}-{4\choose 3})-{5\choose 3}-{3\choose 3}=2017$

And that is the answer. I'm sorry if this was hard to read, I'm not English and I'm not good at writing down my thoughts.

Mathematica

Length@Select[Subsets[Range@25,{3}],GCD@@#==1&]

returns 2017