We don't know what you're asking, can you give an example?

Meanwhile the question you don't seem to have intended turns out to be interesting: counting the coprime pairs. I worked out the answer using the

inclusion-exclusion principle and got 6231 coprime pairs. Then I wrote a Python program and got 6262 coprime pairs. So I'm close and have an error either in my combinatorics or my code. Getting late so I'll have to wait till tomorrow to resolve the discrepancy.

What I did was to count the number of noncoprime pairs and subtract that from 10,000.

For example if both members of a pair are divisible by 2 that's a noncoprime. For convenience I say "2 divides the pair."

There are 50 even numbers between 1 and 100, computed as 100/2 = 50. All divisions are integer divisions throughout. The number of such pairs is 50^2 = 2500.

Likewise there are 100/3 = 33^2 = 1089 pairs divisible by 3.

But now I've counted pairs like (6,6) twice. I have to subtract off (100/6)^2. Continuing like this I have to subtract off all the pairwise intersections. But if a pair is in a three-fold intersection I've overcounted the subtractions and have to add that one back. This can get confusing but the inclusion-exclusion principle lets you just crank it out using a formula. You take the sum of all the odd-fold intersections (including the original sets of pairs divisible by 2, 3, 5, and 7); minus the sum of all the even-fold pairs.

Gotta go find my 31 pair discrepancy.

(Edit) -- I've got it, small arithmetic mistake. 6262 coprime pairs, computed two different ways, by inclusion-exclusion and by program.