If we consider 4 variables i,j,k,l.How many configurations of $i=j\neq k\neq l$,$i=j=k=l$ etc are possible with $= $ or/and $\neq$? I have tried taking combinations of 1 $=$,2 $=$'s and 3 $=$'s..but the required answer which is 15 is not coming.
 A: The cases in which 4, 3, or none of them equal are easy, using combinations we just have $C(4,4)=1$, $C(4,3)=4$, and $C(4,0)=1$.  The cases in which 2 are equal are slightly trickier, as we could have $i=j\neq k\neq l$ or $i=j\neq k=l$.  The former case is once again easy, $C(4,2)=6$.  The latter case requires a bit of thinking.  Obviously, there are $C(4,2)*C(2,2)=6$ ways to do this, but half of these are going to be duplicates (i.e. $i=j\neq k=l$ is the same as $k=l\neq i=j$).  So divide the result by 2: $C(4,2)*C(2,2)/2=3$.
Summing them, we get $1+4+6+3+1=15$.
A: If you can distinguish each variable (i.e., $i=j, k=l, i \neq k$ and $i=k, j=l, i \neq j $ are different cases, for example) then I think the other answers have overlooked a simpler approach. 
You can take equivalence classes on the set $X=\{i, j, k, l\} $ which gives you a partition of the set, but the total number of partitions is $B_n$ the $n$-th Bell number (in this case $n=4$ and $B_n=15$)
A: In the set of variables, $\{i,j,k,l\}$, could appear one, two , three, or four distinct values.
There is only one way for all four variables to have the same value.  Symmetrically, there is only one way for all four varialbes to have different values.  \begin{align}
  i &= j = k = l  \text{ and}  \\
  i &\neq j, i \neq k, i \neq l, j \neq k, j \neq l, k \neq l  \text{, respectively.}
\end{align}
For two values, either one variable has one value and three variables the other, or two have one and two the other.  In the first case, we may choose which of the four variables is unequal to the other three, giving $4$ possibilities.  In the second case, once we choose two variables that are equal, we force the other two to be equal and there are $\dfrac{4 \cdot 3}{2 \cdot 2} = 3$ ways to choose two variables from four (since we do not care about the order of choices or which pair is which).  \begin{align}
    i &= j = k, i \neq l, j \neq l, k \neq l  \text{;}  \\
    i &= j = l, i \neq k, j \neq k, k \neq l  \text{;}  \\
    i &= k = l, i \neq j, k \neq k, j \neq l  \text{; and}  \\
    j &= k = l, i \neq j, i \neq k, i \neq l  \text{.}  \\
    i &= j, k = l, i \neq j, i \neq l, j \neq k, j \neq l  \text{;}  \\
    i &= k, j = l, i \neq j, i \neq l, j \neq k, k \neq l  \text{; and}  \\
    i &= l, j = k, i \neq j, i \neq k, j \neq l, k \neq l  \text{.}
\end{align}
For three values, exactly two variables are equal and there are $\dfrac{4 \cdot 3}{2} = 6$ ways to choose those two variables.  \begin{align}
    i &= j, i \neq k, i \neq l, j \neq k, j \neq l, k \neq l  \text{;}  \\
    i &= k, i \neq j, i \neq l, j \neq k, k \neq l, j \neq l  \text{;}  \\
    i &= l, i \neq j, i \neq k, j \neq l, k \neq l, j \neq k  \text{;}  \\
    j &= k, i \neq j, j \neq l, i \neq k, k \neq l, i \neq l  \text{;}  \\
    j &= l, i \neq j, i \neq l, j \neq k, k \neq l, i \neq k  \text{; and}  \\
    k &= l, 1 \neq k, i \neq l, j \neq k, j \neq l, i \neq j  \text{.}
\end{align}
Putting these together, we get $15$ consistent assignments.
