In $A_{3}A_{2}A_{1}A_{0}$ there are maximum 16 terms.
Similarly in$B_{3}B_{2}B_{1}B_{0}$ also has 16 terms.
So, for $A>B$
When $B=0$, A could take 1 to 15, i.e. 15 terms.
When $B=1$, A could take 14 terms.
..................................
When $B=14$, A could take only 15
So, 15 terms of B could takes $1+2+3+..........+15$ terms for A
So, total $\frac{15\times 16}{2}$=120 terms