I didn't receive an answer on Mathematics S.E. at the time of posting. So I cross posted here.
How can I find the number of k-permutations of $r_1, r_2, r_3, \cdots , r_x$ objects? $r_x$ implies that there are x types of objects.
Here is an example with $k = 15, r_1 = 4, r_2 = 5, r_3 = 8, r_4 = 3$.
I have 20 letters from the alphabet. There are some duplicates - 4 of them are a, 5 of them are b, 8 of them are c, and 3 are d. How many unique 15-letter permutations can I make?
Here's a sketch of an idea. Consider first the problem of computing the number of permutations of $k$ elements given the number of elements $n_i$ in each class. This is easy: it works out to $$\frac{k!}{\Pi_i n_i !}$$ because in any fixed string, you can permute the elements in a class without changing the string.
Now consider the problem of writing down the different partitions of $k$ into the sum of $x$ labelled numbers. Merely finding the sum itself can be done by determining the coefficient of $y^k$ in the polynomial $$ \Pi_i \frac{y^{r_i+1}-1}{y-1}$$
But this is not enough, since you need to weight each such partition by a different number. However, we know that the contribution of $y^i$ must be $1/i!$, so the overall number we are looking for is $k!$ times the coefficient of $y^k$ in the polynomial $$ \Pi_i \sum_{j=0}^{r_i} \frac{y^j}{j!}$$
To make your question more readable, I define variables with the same letter as each mathematical object's first letter. Your question becomes : how to find $p$ permutations of $t_1, t_2, \cdots, t_{k - 1}, t_k$ things. $t_k$ entails $k$ kinds of $t$ things.
To deduce the formula for all the unique permutations of length $\color{red}{L}$ of $\{t_1,t_2,...,t_k\}$, we must find all combinations $=\{c_1,c_2,...,c_k\}$ where $0 \leq c_k \leq t_k$, such that $ \color{red}{c_1 + \cdots + c_k = L} $.
What we need, is the product of the factorials of the elements of that combination: $ \color{limegreen}{\prod_{i=1}^k c_i!} $
Define the number of combinations as $J$. In closing, to answer your question, the number of permutations is $$ = \sum_{j=1}^J \frac{\color{red}{L}!}{\color{limegreen}{\prod_{i=1}^k c_i!}} = \sum_{\color{red}{c_1 + \cdots + c_n = L}} \binom{\color{red}{L}}{c_1,c_2, \cdots, c_n}, $$
as a closed form expression with a Multinomial Coefficient. QED.
Originally, OP worded this question too verbosely as follows.
How can I find the number of k-permutations of n objects, where there are x types of objects, and $r_1, r_2, r_3 ... r_x$ give the number of each type of object?
But it's pleonastic to moot $n$ at all, because $n \stackrel{\text{def}}{=} r_1 + r_2 + r_3 + \cdots + r_x.$
"where there are x types of objects" is redundant too! $r_x$ itself entails $x$ types, because by definition, $r_x$ means $x$ types of $r$ objects.
