Let $T_n$ denote the set of $n$-tuples $\left(b_1, \ldots, b_n \right)$ of non-negative integers such that $$\sum_{i=1}^{n}ib_i=n.$$ I am trying to simplify the sum (whose indicies depend solely on $n, j$, and $k$)
\begin{align*} \sum_{\underset{b_{j}=k}{\left(b_{i}\right)_{i\le n}\in T_{n}}}\frac{n\left(n-1\right)\cdots\left(n-\sum_{i\le n}b_{i}+2\right)}{\prod_{l\le n}\left(b_l !\right)}. \end{align*}
where one of the components (component $b_j$) is equal to some number $k$ specified in advance.
How can I sum over all tuples in $T_n$ such that the $j$th component is fixed?
I am seeking a symbolic solution for general $n, k$, and $j$. In many cases, the sum will be zero (e.g., the largest $b_n$ can be is $1$ due to the weighted sum constraint on $T_n$.
Example For example, if $n=5$ and the second component is fixed to be $1$ ($b_2 = 1)$, then the sum above has two terms corresponding to the vectors $(3,1,0,0,0)$ and $(0,1,1,0, 0)$.