Number of permutations of r items from T total items of which n are distinct. Lets say I have T total items and out of them n are distinct. So n <= T.
In other words we can have m1 items of item 1 and m2 items of item 2....and so on, and that $m_1 + m_2 +\cdots +m_n = T$.
What are the number of permutations of r items taken out of those T items where r <= T? 
If r == T, the answer is : $T!/( m_1! m_2! \cdots m_n!)$
Question is : What if r < T ? What are the number of permutations?
Note I am not looking for number of combinations where order doesn't matter, but number of permutations where order does matter.
I know the combination of r items from n distinct items is C(n+r-1,r).
 A: Consider the the following expansion 
$$
\eqalign{
  & \left( {x_{\,1}  + x_{\,2}  +  \cdots  + x_{\,n} } \right)^{\,2}  =   \cr 
  & \left( {x_{\,1}  + x_{\,2}  +  \cdots  + x_{\,n} } \right)\left( {x_{\,1}  + x_{\,2}  +  \cdots  + x_{\,n} } \right) =   \cr 
  &  = x_{\,1} ^{\,2}  + x_{\,1} ^{\,1} x_{\,2} ^{\,1}  + x_{\,2} ^{\,1} x_{\,1} ^{\,1}  + x_{\,2} ^{\,2}  +  \cdots  =   \cr 
  &  = \sum\limits_{\left\{ {\matrix{   {0\, \le \,j_{\,k} \left( { \le \,2} \right)}  \cr    {j_{\,1}  + \,j_{\,2}  + \, \cdots  + \,j_{\,n} \, = \,2}  \cr  } } \right.\;}
 {\left( \matrix{  2 \cr  j_{\,1} ,\,j_{\,2} ,\, \cdots ,\,j_{\,n}  \cr}  \right)\;x_{\,1} ^{j_{\,1} } \;x_{\,2} ^{j_{\,2} } \; \cdots \;x_{\,n} ^{j_{\,n} } }  \cr} 
$$
we can see that it counts either $x_1x_2$ and $x_2x_1$, i.e. order is taken into account
Here, each item can appear from $0$ to $2$ times.
You are instead looking for
$$
\eqalign{
  & \left( {x_{\,1}  + x_{\,2}  +  \cdots  + x_{\,n} } \right)^{\,r} \quad \left| {\;rep.\,x_{\,k}  \le m_{\,k} } \right.\quad  =   \cr 
  &  = \sum\limits_{\left\{ {\matrix{
   {0\, \le \,j_{\,k}  \le \,m_{\,k} }  \cr 
   {j_{\,1}  + \,j_{\,2}  + \, \cdots  + \,j_{\,n} \, = \,r}  \cr 
 } } \right.\;} {\left( \matrix{
  r \cr 
  j_{\,1} ,\,j_{\,2} ,\, \cdots ,\,j_{\,n}  \cr}  \right)\;x_{\,1} ^{j_{\,1} } \;x_{\,2} ^{j_{\,2} } \; \cdots \;x_{\,n} ^{j_{\,n} } }  \cr} 
$$
where each index is limited to a given max repetition $m_k$., and thus for the corresponding number
$$ \bbox[lightyellow] {  
N(r,n,{\bf m}) = \sum\limits_{\left\{ {\matrix{
   {0\, \le \,j_{\,k}  \le \,m_{\,k} }  \cr 
   {j_{\,1}  + \,j_{\,2}  + \, \cdots  + \,j_{\,n} \, = \,r}  \cr 
 } } \right.\;} {\left( \matrix{
  r \cr 
  j_{\,1} ,\,j_{\,2} ,\, \cdots ,\,j_{\,n}  \cr}  \right)} 
 } \tag{1}$$
The multinomial in fact, counts the number of permutations of $j_1$ objects of type 1 and $j_2$ objects of type 2 and etc.
In standard combinatorics jargon we are dealing with the number of   

words of length $r$, from alphabet $\{1,2, \cdots , n \}$, with allowed repetitions of each character up to $m_k \; |\, 1 \le k \le n$.

Identity (1) can be recasted in various ways, which will result more or less appealing depending on the characteristics
of the vector $\bf m$.
Let's note first of all, that wlog $\bf m$ can be permuted , so arranged in (e.g.) non decreasing order.
Let's examine some particular cases.
a) $m_k=1 \quad \to$ words with different characters
Calling $\bf u$ the vector with all components at 1, 
$$
\begin{array}{l}
 N(r,n,{\bf u}) = \sum\limits_{\left\{ {\begin{array}{*{20}c}   {0\, \le \,j_{\,k}  \le \,1}  \\   {j_{\,1}  + \,j_{\,2}  + \, \cdots  + \,j_{\,n} \, = \,r}  \\
\end{array}} \right.\;} {\left( \begin{array}{c} r \\  j_{\,1} ,\,j_{\,2} ,\, \cdots ,\,j_{\,n}  \\ 
 \end{array} \right)}  =  \\ 
  = \left( \begin{array}{c} n \\  r \\ 
 \end{array} \right)\sum\limits_{\left\{ {\begin{array}{*{20}c}
   {j_{\,1}  = j_{\,2}  =  \cdots  = j_{\,r}  = \,1}  \\
   {j_{\,r + 1}  = j_{\,r + 2}  =  \cdots  = j_{\,n}  = \,0}  \\
   {j_{\,1}  + \,j_{\,2}  + \, \cdots  + \,j_{\,r} \, = \,r}  \\
\end{array}} \right.\;}
 {\left( \begin{array}{c} r \\  j_{\,1} ,\,j_{\,2} ,\, \cdots ,\,j_{\,n}  \\ 
 \end{array} \right)}  =  \\ 
  = \left( \begin{array}{c} n \\  r \\ 
 \end{array} \right)\frac{{r!}}{{1! \cdots 1!0! \cdots 0!}} = n^{\,\underline {\,r\,} }  \\ 
 \end{array}
$$
which is obvious : $n^{\,\underline {\,r\,} } $ denotes the Falling Factorial.
b) $r \le m_k \quad \to$ general words
$$
N(r,\,n,\,r\,{\bf u} \le {\bf m}) = \sum\limits_{\left\{ {\begin{array}{*{20}c}
   {0\, \le \,j_{\,k} \left( { \le \,r} \right)}  \\
   {j_{\,1}  + \,j_{\,2}  + \, \cdots  + \,j_{\,n} \, = \,r}  \\
\end{array}} \right.\;} {\left( \begin{array}{c}   r \\  j_{\,1} ,\,j_{\,2} ,\, \cdots ,\,j_{\,n}  \\ 
 \end{array} \right)}  = n^{\,r} 
$$
which is obvious as well.
Therefore in general we will have
$$
n^{\,\underline {\,r\,} }  \le N(r,\,n,\,{\bf m}) \le n^{\,r} 
$$
c) composition of words
Let's split the alphabet into two, one with $q$ characters and the other with $n-q$ characters.
We divide the vector $\bf m$ in two, accordingly. We may well put at 0 the max repetition of the characters not included.
A word of $r$ characters may be formed by $s$ characters of the first alphabete and $r-s$ of the second one, with $0 \le s \le r$.
We may consider the words from the total alphabet as made by words from the first interspersed with words from the second.
The $s$ characters are dispersed into the $r$ places, maintaining their order in $\binom{r}{s}$ ways.
We obtain in fact the interesting relation
$$ \bbox[lightyellow] {  
\begin{array}{l}
 N(r,\,n,\,{\bf m}_{\,q}   \oplus {\bf m}_{\,n - q} ) = \sum\limits_{\left\{ {\begin{array}{*{20}c}
   {0\, \le \,j_{\,k}  \le \,m_{\,k} \;\left| {\;1\, \le \,k\, \le \,n} \right.}  \\
   {j_{\,1}  + \,j_{\,2}  + \, \cdots  + \,j_{\,n} \, = \,r}  \\
\end{array}} \right.\;} {
\left( \begin{array}{c} r \\  j_{\,1} ,\,j_{\,2} ,\, \cdots ,\,j_{\,n}  \\  \end{array} \right)}  =  \\ 
  = \sum\limits_s {\sum\limits_{\left\{ {\begin{array}{*{20}c}
   {0\, \le \,j_{\,k}  \le \,m_{\,k} \;\left| {\;1\, \le \,k\, \le \,q} \right.}  \\
   {j_{\,1}  + \,j_{\,2}  + \, \cdots  + \,j_{\,q} \, = \,s}  \\
   {0\, \le \,j_{\,k}  \le \,m_{\,k} \;\left| {\;q + 1\, \le \,k\, \le \,n} \right.}  \\
   {j_{\,q + 1}  + \,j_{\,q + 2}  + \, \cdots  + \,j_{\,n} \, = \,r - s}  \\
\end{array}} \right.\;} {\frac{{r!}}{{s!\left( {r - s} \right)!}}\frac{{s!}}{{j_{\,1} !\,\, \cdots j_{\,q} !}}\frac{{\left( {r - s} \right)!}}{{j_{\,q + 1} !\,\, \cdots j_{\,n} !}}} }  =  \\ 
  = \sum\limits_s {\left( \begin{array}{c} r \\  s \\  \end{array} \right)
\sum\limits_{\left\{ {\begin{array}{*{20}c}
   {0\, \le \,j_{\,k}  \le \,m_{\,k} \;\left| {\;1\, \le \,k\, \le \,q} \right.}  \\
   {j_{\,1}  + \,j_{\,2}  + \, \cdots  + \,j_{\,q} \, = \,s}  \\
\end{array}} \right.\;} {\frac{{s!}}{{j_{\,1} !\,\, \cdots j_{\,q} !}}} } \;\sum\limits_{\left\{ {\begin{array}{*{20}c}
   {0\, \le \,j_{\,k}  \le \,m_{\,k} \;\left| {\;q + 1\, \le \,k\, \le \,n} \right.}  \\
   {j_{\,q + 1}  + \,j_{\,q + 2}  + \, \cdots  + \,j_{\,n} \, = \,r - s}  \\
\end{array}} \right.\;} {\frac{{\left( {r - s} \right)!}}{{j_{\,q + 1} !\,\, \cdots j_{\,n} !}}}  =  \\ 
  = \sum\limits_{0\, \le \,s\left( { \le \,r} \right)} {\left( \begin{array}{c} r \\  s \\ 
 \end{array} \right)N(s,\,q,\,{\bf m}_{\,q} )N(r - s,\,n - q,\,{\bf m}_{\,n - q} )}  =\\ 
  = \sum\limits_{0\, \le \,s\left( { \le \,r} \right)} {\left( \begin{array}{c} r \\  s \\ 
 \end{array} \right)N(s,\,n,\,{\bf m}_{\,q} )N(r - s,\,n,\,{\bf m}_{\,n - q} )}  =\\ 
 \end{array}
 } \tag{2}$$
where the last two lines comes from understanding, equivalently:
 - the vectors ${\bf m}_{\,q}$ and ${\bf m}_{\,n-q}$ as two vectors with the dimensions indicated
and then $\oplus$ meaning "join";
 -  ${\bf m}_{\,q}$ and ${\bf m}_{\,n-q}$ as vectors of dimension $n$ with complementary null components
and then $\oplus$ meaning effectively "plus".   
That says that $N$ is given by the binomial convolution of its parts, same as it is
for its two limit  expressions
$$
n^{\,r}  = \sum\limits_{0\, \le \,s} {\binom{r}{s} q^{\,s} \left( {n - q} \right)^{\,r - s} } 
\quad n^{\,\underline {\,r\,} }  = \sum\limits_{0\, \le \,s} {\binom{r}{s}q^{\,\underline {\,s\,} } \left( {n - q} \right)^{\,\underline {\,r - s\,} } } 
$$

So you are right, a recursive approach is well viable.

