You’re not using the terms permutation and combination with their usual meanings. If I understand you correctly, you have $n$ kinds of object (e.g., the numbers $0,1,\dots,n-1$), and you want to form strings of $k$ of them, allowing repetitions and not requiring that every kind of object appear. (In your example, $n=4$ and $k=2$.) You want to count such strings in two ways.
(1) What you’re calling permutations: the order of the $k$ things matters. This is the case that Anthony Labarre answered: there are $n$ ways to choose the first object, $n$ ways to choose the second, and so on right down to the $k$-th and last object, so there are altogether $n^k$ such strings. If the objects are the numbers $0,1,\dots,n-1$, you’re counting the number of base $n$ integers that can be written with at most $k$ digits.
(2) What you’re calling combinations: all you care about now is how many of each kind of object appears in the string. In your example, for instance, $0111,1011,1101$, and $1110$ all have one $0$ and three $1$s, so they get counted as a single ‘combination’. Suppose that the objects are the numbers $0,1,\dots,n-1$. For $i=1,2,\dots,n-1$ let $k_i$ be the number of times $i$ occurs in a string; since each string has $k$ objects altogether, we must have $k_0 + k_1 + \dots + k_{n-1} = k$, where each $k_i$ is a non-negative integer. Conversely, any list $\langle k_0,k_1,\dots,k_{n-1} \rangle$ in which $k_0 + k_1 + \dots + k_{n-1} = k$ represents one possible ‘combination’. Thus, there are as many ‘combinations’ as there are solutions to the equation $k_0 + k_1 + \dots + k_{n-1} = k$ in non-negative integers. This is a standard problem; the answer is ${{n+k-1} \choose k}$. The discussions here and here may be useful.
Letting $i$ correspond to the first digit, $j$ to the second, $k$ to the third and $l$ to the fourth, the number of numbers is $$\sum_{l=0}^9 \sum_{k=0}^l \sum_{j=0}^k \sum_{i=0}^j 1$$ $$= \sum_{l=0}^9 \sum_{k=0}^l \sum_{j=0}^k \binom{j+1}{1}$$ $$= \sum_{l=0}^9 \sum_{k=0}^l \binom{k+2}{2}$$ $$= \sum_{l=0}^9 \binom{l+3}{3}$$ $$= \binom{9+4}{4} = 715$$
Best Answer
This is very closed to De Bruijn sequences.
Let's call $A$ your alphabet (for decimal numbers $A=\{0,\ldots,9\}$) and $n$ the length of the contained sequences. We look for sequences containing all elements as (possibly) cyclic substring.
Consider the digraph $G = (V,E)$, where:
$$V=A^{n-1}$$ $$E=\{((u_1, \ldots, u_{n-1}), (u_2, \ldots, u_{n-1}, v)), (u_1,\ldots, u_{n-1},v \in A^n\}$$
Then sequences containing all elements of $A^n$ as substring are exactly walks in $G$ going through every edge.
Each node has an in and out degree of $|A|$, so this graph has an Eulerian cycle, which leads to a sequence of length $|A|^n$ minimal possible.
If we delete the cycle, the same graph still works. However, the minimal length we get is now the cost of writing the $n-1$ first letters plus the $|E|$ edges, i.e. $n-1+|A|^n$. For $n=4$, $A=\{0,\ldots,9\}$, we get $10003$.