Complex conjugate Young Tableaux representation I have been studying Young Tableaux representation from youtube to represent $2\times 2$ and other examples to in $SU(n)$ symmetry. But i am unable to understand nor able to find relevant answers of how to represent conjugate i.e how can one write $\bar{3}$ in a square box in $SU(3)$ representation is actually 2 and $\bar{6}$ in $SU(3)$ is represented as 3 in $SU(3)$. By saying 3 i mean i sy its dimension. 
 A: The confusion here is that conjugating a Young diagram does conjugate the representation.  This is compounded by the regrettable use of the dimension to label the representation.  
In the specific case of $\mathfrak{su}(3)$ (and more generally for $\mathfrak{su}(n)$), a better solution is to use the Dynkin labels $(a,b)$.  
For $\mathfrak{su}(3)$, an irrep with Dynkin label $(a,b)$ corresponds to a two-rowed Young diagram with $a+b$ boxes on the first row, and $b$ boxes on the second row: $\{a+b,b\}$.  The advantage of the Dynkin labels is that the representation conjugate to $(a,b)$ is just $(b,a)$.  It will have the same dimension as $(a,b)$ but does not correspond to the diagram conjugate $\{a+b,b\}$.  Indeed it's not hard to see that the diagram conjugate to $\{a+b,b\}$ will have in $a+b$ rows, so it cannot describe an $\mathfrak{su}(3)$ irrep if $a+b>3$.  The irrep $(b,a)$ is associated with the diagram $\{b+a,a\}$.  Note that the irreps $(b,a)$ is associated with a Young diagram having a different number of boxes than its conjugate irrep $(a,b)$.
There are standard formulas to obtain the dimension of the representation in terms of Dynkin labels:  for $\mathfrak{su}(3)$ we have
\begin{align}
\hbox{Dim}(a,b)=\frac{1}{2}(a+1)(a+b+2)(b+1) \tag{1}
\end{align} 
so that, for the irrep corresponding to a single box, $(a,b)=(1,0)$, with Young diagram $\{1\}$ (a single box) and dimension $(2\times 3\times 1)/2=3$.  This is what you call $\textbf{3}$.
The conjugate irrep is $(0,1)$, which you call $\bar{\textbf{3}}$, and  is associated to Young diagram $\{1,1\}$, i.e. two boxes in a single column.  Its dimension is also $3$ since the dimensionality formula (1) is symmetric under the interchange of $a$ and $b$. 
