What is the chance that the balls are evenly distributed in the box? Six different balls are kept into 3 different boxes randomly so that no box being empty.What is the chance that the balls are evenly distributed in the box?
I have learnt one formula that number of ways in which $m\times n$ different objects can be distributed equally among $n$ persons$=\frac{(mn)!}{(m!)^n}$
So the number of ways in which 6 different objects can be distributed equally among 3 boxes$=\frac{6!}{(2!)^3}$
I counted the total number of ways as$\binom{6-1}{3-1}=10$
But this is giving me the wrong answer,correct answer is $\frac{1}{6}$.What mistake have i done?
Is evenly distributed not same as the equally distributed?
Please help me. 
 A: $\dfrac{6!}{2^3}$ is correct for favorable ways, 
But total ways of putting distinct balls into distinct boxes would be $3^6$, as each ball has $3$ choices for box. 
[edit:]
However, as we also need to restrict this so that no box is empty, we use the Principle of Inclusion and Exclusion (PIE) to exclude when at least 1 box is empty.   So that is $3^6-3\cdot 2^6 +3$
$$\dfrac{6!}{2^3(3^6-3\cdot 2^6+3)} = \dfrac 1 6$$
A: The balls are numbered from one to six, and each of them went into one of the boxes $A$, $B$, $C$ with equal probability, and independently. A priori there would be $3^6$ equiprobable cases, but we are told that none of the boxes remained empty. In any case the result of the experiment can be encoded as a word of length $6$ over the alphabet $\{A,B,C\}$.
Cardinalitywise there  are  three different types of admissible distributions of the six balls, namely $$T_1:\ 4+1+1, \qquad T_2: \ 3+2+1, \qquad T_3: \  2+2+2\ .$$
We now have to count the number of words for each of the three types. For type $T_1$ we can choose the letter occurring four times in $3$ ways; then we can place the four equal letters in ${6\choose 4}$ ways, and finally we can allocate the two remaining letters in $2$ ways, making a total of $90$ words of type $T_1$. For $T_2$ we can choose the letter occurring three times in $3$ ways and the letter occurring two times in $2$ ways. Then we can place these letters in ${6!\over 3!\>2!\>1!}$ ways, making a total of $360$ words of type $T_2$. For $T_3$ we can place $A$ in ${6\choose2}$ ways, then $B$ in ${4\choose 2}$ ways, and the two last slots  are for $C$. This amounts to $90$ words of type $T_3$. 
The probability in question is then given by
$${|T_3|\over|T_1|+|T_2|+|T_3|}={90\over90+360+90}={1\over6}\ .$$
