Distinct numbers with two $3$s and two $7$s: $\binom{4}{2}=6$.
Distinct numbers with two $3$s and one or fewer $7$s: $\binom{4}{2}3\cdot2=36$.
Distinct numbers with two $7$s and one or fewer $3$s: $\binom{4}{2}3\cdot2=36$.
Distinct numbers with one or fewer $7$s and one or fewer $3$s: $4\cdot3\cdot2\cdot1=24$.
Total: $6+36+36+24=102$
With larger alphabets,
Suppose there are $a$ numbers with 4 or more in the list, $b$ numbers with exactly 3 in the list, $c$ numbers with exactly 2 in the list, and $d$ numbers with exactly 1 in the list.
Distinct numbers with all 4 digits the same: $a$
Distinct numbers with 3 digits the same: $\binom{4}{3}(a+b)(a+b+c+d-1)$
Distinct numbers with 2 pairs of digits: $\binom{4}{2}\binom{a+b+c}{2}$
Distinct numbers with exactly 1 pair of digits: $\binom{4}{2}(a+b+c)\binom{a+b+c+d-1}{2}2!$
Distinct numbers with no pair of digits: $\binom{a+b+c+d}{4}4!$
Total: $a+4(a+b)(a+b+c+d-1)+6\binom{a+b+c}{2}+12(a+b+c)\binom{a+b+c+d-1}{2}+24\binom{a+b+c+d}{4}$
Apply to the previous case: $a=b=0$, $c=2$, and $d=2$:
$0+0+6\binom{2}{2}+12(2)\binom{3}{2}+24\binom{4}{4}=102$