Probability question and how to approach it 
Nine tiles are numbered 1, 2, 3, . . . , 9. Each of three players randomly selects and keeps three of the tiles, and sums those three values. Find the probability that all three players obtain an odd sum.

There are 5 odd numbers from 1 to 9, to have 3 of them all get odd sums, that requires a 1-1-3 split for the 5 numbers:
1st approach: person A picks 3 cards, person B picks 3 out of the remaining 6. Brute force calculation
2nd approach: only consider the 5 odd numbers, and ways to place them in three groups so that they are in 1-1-3. ( I don't think this is correct )
3rd approach: There are three groups, 9 numbers are randomly ordered so three numbers land in each group.
Are those correct? and is 1st and 3rd way of interpreting the question going to give me the same result?
Thank you!
 A: Yes, the first and third are correct approaches and will give the same result (if I understand you correctly).  The second approach is not quite right, but can be modified to be correct, and the modified approach is cleaner than your first and third approaches.
To see that the second approach is wrong, notice that for an arrangement of odd numbers into groups of size $3, 1,$ and $1$, the even numbers can then be distributed in ${4 \choose 2} = 6$ ways.  However, if the odd numbers are arranged into groups of size $3, 2, 0$, then the even numbers can be distributed in ${4 \choose 1} = 4$ ways.  Thus not all arrangements of just the odd numbers are equally likely.
A cleaner approach
To avoid the problems of the second approach, consider a list of $9$ slots rather than three groups of three.
You are distributing $5$ odd numbers and $4$ even numbers into the list of $9$ slots.
But notice that we don't care which even number goes where; an even number is an even number regardless of where it is.  Similarly for odd numbers.  So we might as well call all the odd numbers "$1$" and all the even numbers "$0$".
Under this view, the problem becomes

How many ways can $5$ ones and $4$ zeros be placed into a list of $9$ slots so that each group of three contains either $3$ or $1$ ones?

and should be much nicer to solve.
