If $2$ variables a and b can form 3 combinations such as $a>b$, $aHow many unique combinations can $4$ variables have when compared together
For example :
$$a=b>c<d$$
I tried solving it and came to a conclusion that
if $4$ variables can have $6$ unique combination when grouped in a pair and each variable can have $3$ different combination then $4$ variables should have $18$ unique combinations
 A: Unclear if I am misinterpreting the question. 
Anyway, my take is:
Let $S_k$ denote the computation for Case $k$.
$\underline{\text{Case 1 : All 4 variables distinct}}$
The $4$ variables can be permuted in $4!$ ways. 
$$S_1 = 24.$$

$\underline{\text{Case 2 : One equal pair, the other 2 variables distinct}}$
There are $\binom{4}{2} = 6$ ways of choosing which two variables are equal.
Once this is done, you (in effect) have $3$ units, which can be permuted in $3! = 6$ ways.
$$S_2 = 6 \times 6 = 36.$$

$\underline{\text{Case 3 : Two equal pairs}}$
There are $3$ ways of determining how the two pairs should be formed.
Once this is done, you (in effect) have $2$ units, which can be permuted in $2! = 2$ ways.
$$S_3 = 3 \times 2 = 6.$$

$\underline{\text{Case 4 : A triplet}}$
There are $4$ ways of determining which variable is the distinct variable.
Once this is done, you (in effect) have $2$ units, which can be permuted in $2! = 2$ ways.
$$S_4 = 4 \times 2 = 8.$$

$\underline{\text{Case 5 : A quartet}}$
There is only $1$ way of forming such a quartet.
$$S_5 = 1.$$
$\underline{\text{Final Computation}}$
$$S_1 + S_2 + \cdots + S_5 = 24 + 36 + 6 + 8 + 1 = 75.$$
