Choosing three distinct numbers giving even sum. In how many ways can we choose three distinct numbers from the set $\{1,2,...,99\}$ so that their sum is even?
I found a solution in this StackExchange post .
My question is:
When working with the case with two odd numbers and one even number, I first use the principle of counting to find all sequences containing two odd numbers and one even number as $50\times49\times49$. Then I divide by $3!$ as different permutations of the sequence don't count. I get a fractional value. Instead, if I divide by $2!$, i.e., only stripping away the arrangements of the two odd numbers (not all three digits!), I get the right answer.
I have some difficulties understanding this problem:


*

*$\binom{50}{2}\times\binom{49}{1}$ implies we are considering  arrangements of 2-subsets of the set of odd numbers and 1-subsets of the set of even numbers. I don't understand why we don't divide this by $2!$ to discount the permutations of these two subsets. 

*Why is dividing by $3!$ incorrect?
 A: *

*You can assume that you're choosing the odd numbers first, then the even numbers. Then, since the order of these subsets is fixed, there is no need to divide by $2!$. If you don't fix the order this way, then the counting gets repetitive: you would have to account for different ways of choosing the numbers (odd-even-odd, etc.), then indeed divide at the end to compensate.

*Dividing by $3!$ is incorrect because you're pulling the odd and even numbers from different sets. If your numbers are $13, 48$, and $67$, you could have chosen $13$ as the first odd number, then $67$, or the other way around. There are only two orders for those, just as there is only one way you could have chosen $48$. (You couldn't have chosen $48$ as the first odd number, for example.) This is why dividing by $2!$ gives you the right answer.
A: For me, it often helps to think of sample objects which I am counting.
One of your confusions seems to be what $a \times b$ means in a counting formula.  I like to think of it as choosing among $a$ options first, and then, choosing among $b$ options, in that order (if order matters).  A typical object being chosen is a $2$-vector $(A,B)$, where $A$ is one of the $a$ options and $B$ is one of the $b$ options.  In particular a typical object is not  a $2$-set $\{A, B\}$.


*

*For ${50 \choose 2} \times {49 \choose 1}$, a typical object is a $2$-vector $(A,B)$ where $A$ is a $2$-subset of odd numbers and $B$ is a $1$-subset of even number.  Every desired $3$-subset of $2$ odd numbers and $1$ even number can be written in this way in exactly one way.  So no need to divide.  E.g. $\{23, 4, 17\} = (\{23, 17\}, \{4\})$.


*

*In particular ${50 \choose 2} \times {49 \choose 1}$ does not count $2$odd-even "as well as" even-$2$odd.  For that you would need ${50 \choose 2} \times {49 \choose 1} +  {49 \choose 1} \times {50 \choose 2}$... and then of course you need to divide by $2!$.


*For $50 \times 49 \times 49$, first decide which $49$ means what.  The fact that they are the same number is just a coincidence.  Lets say you arrived at that formula thinking the first $49$ is choosing the second odd number, i.e. you're choosing odd-odd-even.  So then a typical object is a $3$-vector $(A,B,C)$ where $A, B$ are distinct odd numbers and $C$ is even.  Any desired $3$-subset can be written in this $3$-vector form in $2$ ways, e.g. $\{23, 4, 17\} = (23,17,4)$ or $(17,23,4)$, and crucially, $(23,17,4) \neq (17,23,4)$, so you divide by $2!$.


*

*If you decide the first $49$ means the even number, i.e. you're choosing odd-even-odd, it's the same deal.  But again, $50 \times 49 \times 49$ does not include odd-odd-even "as well as" odd-even-odd.  For that you would need $50 \times 49 \times 49 + 50 \times 49 \times 49$.  And if you include even-odd-odd you would need $50 \times 49 \times 49 + 50 \times 49 \times 49 + 49 \times 50 \times 49$... and then you would truly need to divide by $3!$.


