an equation with generalized sum In my work on summation I get this relation
$$\sum_{t_1+t_2+\cdots+t_k=m,1t_1+2t_2+\cdots+kt_k=k}\frac{k!}{t_1!t_2!(2!)^{t_2}\cdots t_k!(k!)^{t_k}}=$$
$$=\sum_{t_1\geq 1}\binom{k}{t_1}\sum_{t_2+\cdots+t_k=m-t_1,2t_2+\cdots+kt_k=k-t_1} \frac{(k-t_1)!}{t_2!(2!)^{t_2}\cdots t_k!(k!)^{t_k}}$$
I ask someone to tell me if that is correct because I am not sure
from $$t_1+t_2+\cdots+t_k=m,1t_1+2t_2+\cdots+kt_k=k$$ follow
$$t_2+\cdots+t_k=m-t_1,2t_2+\cdots+kt_k=k-t_1$$
 A: At first  we do not concentrate on the sums, we just consider the two systems of linear equations:

\begin{align*}
t_1+t_2+\cdots+t_k&=m&\text{and}&&t_2+\cdots+t_k&=m-t_1\tag{1}\\
1t_1+2t_2+\cdots+kt_k&=k&&&2t_2+\cdots+kt_k&=k-t_1
\end{align*}
  We see the right hand side is an equivalence transformation which subtracts $t_1$ on each side of the equations and this is a valid transformation. Both represent  the same situation. So, maybe you actually don't need this transformation.

Please note, the first sum has a flaw as none  of the indices $m,k$ and $t_j$ with $1\leq j\leq k$ is specified. So, in order to appropriately specify the sum you have to declare the valid range of the indices.
Here is an example:

Let $m,k\geq 1$ be positive integers. We consider natural numbers $t_j\geq 1, 1\leq j\leq k$ with
  \begin{align*}
\sum_{{t_1+t_2+\cdots+t_k=m}\atop{t_1+2t_2+\cdots+kt_k=k}}\frac{k!}{t_1!t_2!(2!)^{t_2}\cdots t_k!(k!)^{t_k}}
\end{align*}

and here are some equivalent representations

\begin{align*}
\sum_{{t_1+t_2+\cdots+t_k=m}\atop{t_1+2t_2+\cdots+kt_k=k}}&\frac{k!}{t_1!t_2!(2!)^{t_2}\cdots t_k!(k!)^{t_k}}\\
&=\sum_{{t_1+t_2+\cdots+t_k=m}\atop{t_1+2t_2+\cdots+kt_k=k}}\binom{k}{t_1}\frac{(k-t_1)!}{t_1!t_2!(2!)^{t_2}\cdots t_k!(k!)^{t_k}}\\
&=\sum_{t_1\geq 1}\sum_{{t_1+t_2+\cdots+t_k=m}\atop{t_1+2t_2+\cdots+kt_k=k}}\binom{k}{t_1}\frac{(k-t_1)!}{t_1!t_2!(2!)^{t_2}\cdots t_k!(k!)^{t_k}}\\
&=\sum_{t_1\geq 1}\ldots\sum_{t_k\geq 1}\sum_{{t_1+t_2+\cdots+t_k=m}\atop{t_1+2t_2+\cdots+kt_k=k}}\binom{k}{t_1}\frac{(k-t_1)!}{t_1!t_2!(2!)^{t_2}\cdots t_k!(k!)^{t_k}}\\
&=\sum_{{t_j\geq 1}\atop{1\leq j\leq k}}\sum_{{t_1+t_2+\cdots+t_k=m}\atop{t_1+2t_2+\cdots+kt_k=k}}\binom{k}{t_1}\frac{(k-t_1)!}{t_1!t_2!(2!)^{t_2}\cdots t_k!(k!)^{t_k}}\\
&=\sum_{t_1\geq 1}\binom{k}{t_1}\sum_{{t_1+t_2+\cdots+t_k=m}\atop{t_1+2t_2+\cdots+kt_k=k}}\frac{(k-t_1)!}{t_1!t_2!(2!)^{t_2}\cdots t_k!(k!)^{t_k}}\\
&=\sum_{t_1\geq 1}\binom{k}{t_1}(k-t_1)!\sum_{{t_1+t_2+\cdots+t_k=m}\atop{t_1+2t_2+\cdots+kt_k=k}}\frac{1}{t_1!t_2!(2!)^{t_2}\cdots t_k!(k!)^{t_k}}
\end{align*}

Of course the representation depends on your needs, but in each of the cases above it is not necessary, to switch from the LHS to the RHS of (1).
A: $$\sum_{t_1+t_2+\cdots+t_k=m,1t_1+2t_2+\cdots+kt_k=k}\frac{k!}{t_1!t_2!(2!)^{t_2}\cdots t_k!(k!)^{t_k}}=p(m,k)_{[1]}$$ is number of partitions of k-set into m parts that contain at least 1 element,and
$$\sum_{t_2+\cdots+t_k=m,2t_2+\cdots+kt_k=k}\frac{k!}{t_2!(2!)^{t_2}\cdots t_k!(k!)^{t_k}}=p(m,k)_{[2]}$$ is number of partitions of k-set into m parts that contain at least 2 elements. The question is does hold true the following equation.
$$p(m,k)_{[1]}=\sum_{i\geq 0}\binom{k}{i}p(m-i,k-i)_{[2]}$$
