How many ways can we choose three distinct numbers in the set $ {1,2, …, 99 } $ for their sum to be equal?
I found a solution in this article StackExchange.
My question is:
When I work with the case with two odd numbers and an even number, I first use the principle of counting to find all the sequences containing two odd numbers and an even number. $ 50 times49 times49 $. Then I divide by $ 3! $ as different permutations of the sequence do not count. I receive a fractional value. Instead, if I divide by $ 2! $that is, by erasing only the arrangements of the two odd numbers (not all three digits!), I get the correct answer.
I have some difficulties understanding this problem:

$ binom {50} {2} times binom {49} {1} $ implies that we envision arrangements of 2 subsets of the set of odd numbers and a subset of the set of even numbers. I do not understand why we do not divide that by $ 2! $ to rule out the permutations of these two subsets.

Why divide by $ 3! $ Incorrect?