I have to calculate a product of two probabilities, namely hypergeometric distribution $$P_1(t)=\frac{\binom{\frac{D+m}{2}}{t}\binom{\frac{D-m}{2}}{z-t}}{\binom{D}{z}}$$ and binomial distribution $$P_2(s)=\frac{1}{2^{D-z}}\binom{D-z}{s}$$ where $D$, $m$, $z$, and $\mu$ are constants, while $s$ and $t$ are variables. All the constants are natural numbers. $D$, $m$ is even number and $D\ge m$ so that both $\frac{D+m}{2}$ and $\frac{D-m}{2}$ are always non-negative integers. But, I need a following sum condition for two variables, $$ s,t\in \mathbb N $$$$ s+t=\frac{D+y}{2} $$ where y is another even number and $D\ge y$ so that $\frac{D+y}{2}$ are always non-negative integers as well. (Sorry for messy usage of constant names.) It means that, if $\frac{D+y}{2}=10$, I need the sum of products, namely $P_1(0)P_2(10)+P_1(1)P_2(9)+P_1(2)P_2(8)...+P_1(10)P_2(0)$. Thus, final formula would be as followings.
$$\sum_{s+t=\frac{D+y}{2}}P_1(t)P_2(s)$$ $$=\sum_{s+t=\frac{D+y}{2}}\frac{\binom{\frac{D+m}{2}}{t}\binom{\frac{D-m}{2}}{z-t}\binom{D-z}{s}}{2^{D-z}\binom{D}{z}}$$ $$=\frac{1}{2^{D-z}\binom{D}{z}}\sum_{s+t=\frac{D+y}{2}}\binom{\frac{D+m}{2}}{t}\binom{\frac{D-m}{2}}{z-t}\binom{D-z}{s}$$
How can I simplify this sum of triple product expression as simple as possible? Is there any explicit formula with $s$ and $t$ which could be induced by usage of generating function?
Edit. $m$ and $y$ could be a negative even number (-2, -4...), but $|m|,|y|<D$ thus description above still remains same.
Edit2. I realized that I can remove $s$ with $s=\frac{y+D}{2}-t$, thus expression is same as following without any sum condition.
$$\frac{1}{2^{D-z}\binom{D}{z}}\sum_{t=0}^{\frac{y+D}{2}}\binom{\frac{D+m}{2}}{t}\binom{\frac{D-m}{2}}{z-t}\binom{D-z}{\frac{y+D}{2}-t}$$