\begin{equation} B(a,b+1) = \sum_{i=0}^{a} \genfrac(){0pt}{0}{a}{i} B(i,b). \end{equation}
To show this relationship is true we make the following argument.  First assume $B(a,b)$ is the 
number of possible unique partitions when allocating $a$ items and $b$ subsets already exist.  
We are now assuming we have $b+1$ existing subsets before allocating the $a$ items.  That extra subset must be contained in 
exactly one of the following cases:
\begin{itemize}
\itemsep0em
\item None of those $a$ items are added to that subset: there are $B(a,b)$ unique partitions in 
this case
\item One of those $a$ items are added to that subset: there are 
$\genfrac(){0pt}{1}{a}{1} B(a-1,b)$ unique partitions in this case 
\item Two of those $a$ items are added to that subset: there are 
$\genfrac(){0pt}{1}{a}{2} B(a-2,b)$ unique partitions in this case \\
\vdots
\item All $a$ of those items are added to that subset: there are 
$\genfrac(){0pt}{1}{a}{a} B(0,b)$ unique partitions in this case 
\end{itemize}
Recalling that $\genfrac(){0pt}{1}{a}{i} = \genfrac(){0pt}{1}{a}{a-i}$ and summing all the 
possible cases in reverse order, we have that 
$B(a,b+1) = \sum_{i=0}^{a} \genfrac(){0pt}{1}{a}{i} B(i,b)$.


A useful recurrence formula for these numbers are:
\begin{equation}\label{EQ:eBellRecur} B(a+1,b) = b \, B(a,b) + B(a,b+1). \end{equation}
