Question

Nine tiles are numbered $1,2,3,4,5,6,7,8,9$ respectively. Each of the three players $A,B$ and $C$ randomly selects $3$ tiles one after the other without replacement and they sum up those three values as marked on the tiles. The probability that all three players obtain an odd sum is $\frac{m}{n},$ where $m$ and $n$ are relatively prime. The value of $|2m-n|$ is

Answer 1

Number of ways in which $A,B,C$ can be given $3$ tiles
i.e., $3$ equal groups of $9$ tiles and distributed $=n\left(S\right)=\frac{9!\cdot 3!}{(3!{)}^3\cdot 3!}=1680$

Now, we have $1,3,5,7,9\left(\text{odd numbers}\right);2,4,6,8\left(\text{even numbers}\right)$
In order that all the $3$ persons must have an odd total, one person must have all the $3$ tiles marked with odd number and each of the remaining two persons must have $2$ tiles with even marked and $1$ tile with odd marked.
One person can be selected in ${}^3{C}_1$ ways.
$3$ odd tiles out of $5$ can be selected in ${}^5{C}_3$ ways.
The remaining two odd tiles can be selected and distributed in ${}^2{C}_1$ ways.
Remaining $4$ even tiles can be equally distributed in $\frac{4!\cdot 2!}{(2!{)}^2\cdot 2!}$ ways.
$\therefore n\left(E\right)={}^3{C}_1\cdot {}^5{C}_3\cdot {}^2{C}_1\cdot \frac{4!\cdot 2!}{(2!{)}^2\cdot 2!}$
$=3\cdot 10\cdot 2\cdot 6=360$
Probability, $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{360}{1680}=\frac{3}{14}=\frac{m}{n}$
$\therefore |2m-n|=|2\times 3-14|=8$