How many ways are there to choose integers a, b, and c such that 1 ≤ a < b < c ≤ 2024 and
a + b + c = 2027?
Lets start by ignoring all the constraints in the inequality and just ask for ordered triplets of positive integers {a,b,c} so that a+b+c=2027.
This is actually very simple, its just 2026C2 = (2026*2025)/(2*1) = 2051325.
Now to break this number down into triplets with or without repeats.
2027 is not a multiple of 3, so no triplets have all members equal.
There are 1013 unordered triplets with two terms equal; for each value of a 1 to 1013 then {a,a,2027-2a} is an unordered triplet. Multiply by 3 to get 3039 ordered triplets.
That leaves 2051325-3039 = 2048286 unordered triplets with all three terms distinct. Since there are six permutations, then there are 2048286/6 = 341381 ordered triplets with all three terms distinct.
The system of inequalities set in the problem is in fact asking for the number of ordered triplets with all three terms distinct. So the answer is there are 341381 ways to choose integers a, b, and c such that 1 ≤ a < b < c ≤ 2024 and a + b + c = 2027.
Analytic Solution
