# combinatorial – Choose three distinct numbers giving an even sum.

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:

1. $$binom {50} {2} times binom {49} {1}$$ 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.

2. Why divide by $$3!$$ Incorrect?