Given an alphabet with 6 non-distinct integers, how many distinct 4-digit integers are there? 
How many distinct four-digit integers can one make from the digits
  $1$, $3$, $3$, $7$, $7$ and $8$?

I can't really think how to get started with this, the only way I think might work would be to go through all the cases. For instance, two $3$'s and two $7$'s as one case, one $1$, two $3$'s and one $8$ as another. This seems a bit tedious though (especially for a larger alphabet) and so I'm here to ask if there's a better way.
Thanks.

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 $d=2$:
$0+0+6\binom{2}{2}+12(2)\binom{3}{2}+24\binom{4}{4}=102$
