Combinatorics Question: Alphabet of $16$ letters, $8$ slots, arbitrary blanks If I have an alphabet of $16$ characters and $8$ slots that are filled with any combination of characters (no duplicates except blanks), how do I calculate the total number of combinations?
Edit for clarification
(using a smile set of 3 char (1-3) and 4 slots (using x as blank or null)
xxxx is valid
1xxx is valid
xxx1 is valid
123x is valid
12x3 is valid
3xx3 is not valid
1x1x is not valid
321x is valid

I appreciate everyone help I don't have much math background so im struggling to convey the concept with out a proper vocabulary 
 A: OP didn't mention whether order matters or doesn't matter. So here is an answer for either case:
Order Doesn't Matter
You have an alphabet of $16$ characters and you have $8$ slots to be filled. Then the number of combinations is ${16 \choose 8}$. 
However, you are allowed to use blanks. So suppose you use one blank. Then we have $7$ slots to be filled: that's ${16 \choose 7}$. The same goes for two blanks: we have ${16 \choose 6}$.
So the total number of possible combinations will be, by the rule of sum, ${16 \choose 8} + {16 \choose 7} + \ldots + {16 \choose 1}$, or in summation notation: $$\sum_{i=1}^{8} {16 \choose i}$$
Order Matters
Again, you have $8$ slots to be filled. There are ${16 \choose 8}$ ways to choose the elements. Now you have to order them. For $n$ distinct objects, there are $n!$ permutations. As we will always have $8$ objects, the total number of ways to fill $8$ slots is ${16 \choose 8} \cdot 8!$.
Now take the case with $1$ blank. There are ${16 \choose 7}$ ways to choose the elements. Now, notice that the blank space also counts a distinct element. In this case, we have ${16 \choose 7} \cdot 8!$ combinations.
Now take the case with $2$ blanks. We have ${16 \choose 6}$ ways to choose the elements. We have two blank spaces --- they only count as one distinct element. So we have ${16 \choose 6} \cdot 7!$ combinations. Etc.
So the total number of possible combination, again by the sum rule, will be ${16 \choose 8}8! + {16 \choose 7}8! + {16 \choose 6}7! + {16 \choose 5}6! + \ldots + {16 \choose 1}2!$. I suppose an adequate closed-form expression would be: $${16 \choose 8}8! + {\sum_{i=1}^{7}{{16 \choose i}(i+1)!}}$$
A: Order matters. Then for $n$ blanks you need to chose the n blanks in $\binom{8}{n}$ ways and then choose a permutation of the 8-n characters in $\frac{16!}{(8+n)!}$ ways. So the total number of words is $\sum_{n=0}^8 \frac{16!}{(8+n)!}\binom{8}{n}$
