Here's a question from the AIME competition: https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_4
Nine tiles are numbered $1, 2, 3, \cdots, 9,$ respectively. Each of three players randomly selects and keeps three of the tiles, and sums those three values. The probability that all three players obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
It's also been asked before on Math Stack Exchange before here: Probability question and how to approach it
Here's what I did. Let's say the order in which people take tiles does matter. Then my result is$${{3 \binom{5}{3} 2 \binom{4}{2}}\over{3! \binom{9}{3,3,3}}} = {1\over{28}}.$$However, the answer at the given link, where they don't take into consider order (saying it will be the same final answer) is:$${{3 \binom{5}{3} 2 \binom{4}{2}}\over{\binom{9}{3,3,3}}} = {3\over{14}}.$$So my question is, if we take into consideration the order in which people take tiles, then why do we need an extra $3!$ in the numerator (to cancel out the $3!$ in the denominator)? I thought the $3$ I had already took care of that, since only $1$ of the $3$ players will have $3$ odds, which "fixes" the configuration. So how am I mistaken, what am I missing?