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.
This is known as multichoose; explicitly, the number you're after is "$7$ multichoose $3$." The basic theorem to keep in mind is that "$n$ multichoose $k$" equals $[n-1,k],$ where $$[-,-] : \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$$ is the "symmetric binomial coefficient" defined by $$[a,b] = \frac{(a+b)!}{a!b!}.$$
So "$7$ multichoose $3$" equals $$[7-1,3]$$ which equals $$\frac{9!}{6!3!}$$
which equals $$84.$$
Best Answer
What you are looking for is the Stars and Bars method. This method is used to count things where order does not matter and repetition is allowed.
We are choosing $9$ digits from $\{1,2,3,4,5,6\}$
Let me explain how we can get to the answer.
For example, let's say we want to pick the combination $123456123$. This can be "arranged" as such:
$$••|••|••|•|•|•$$
We picked $2$ ones, $2$ twos, $2$ threes, and $1$ of the rest.
Let's look at another possibility: $111222456$ $$•••|•••| |•|•|•$$
We have three ones, three twos, and one $4$, $5$, and $6$.
Look at what these pictures are showing: we have $14$ symbols and we are choosing $5$ of them to be a bar. Or we could look at it as choosing $9$ of them to be a dot (star)
So the answer to our original question is $$\binom{14}{9} = \binom{14}{5} = 2002$$