Multi-binomial theorem and double product expansion The Multi-binomial theorem says that
$$\prod_{i=1}^d(x_{i}+y_{i})^{n_{i}}=(x_{1}+y_{1})^{n_{1}}\dotsm(x_{d}+y_{d})^{n_{d}} = \sum_{k_{1}=0}^{n_{1}}\dotsm\sum_{k_{d}=0}^{n_{d}} \binom{n_{1}}{k_{1}}\, x_{1}^{k_{1}}y_{1}^{n_{1}-k_{1}}\;\dotsc\;\binom{n_{d}}{k_{d}}\, x_{d}^{k_{d}}y_{d}^{n_{d}-k_{d}}.
$$
How could we express
$$
\prod_{i=1}^d\prod_{j=1}^k (x_{ij}+y_{ij})^{n_{ij}},
$$
in the light of the Multi-binomial theorem?
If that is too messy, what about the following simpler expression?
$$
\prod_{i=1}^d\prod_{j=1}^k (1-x_{ij}).
$$
 A: The procedure for expanding the double product is the same as the one for the single product as you did well.
Let us expand the double product by using the binomial theorem:$$\prod_{i=1}^d \prod_{j=1}^k (x_{ij}+y_{ij})^{n_{ij}} = (x_{11} + y_{11})^{n_{11}} \cdots (x_{1k}+y_{1k})^{n_{1k}}(x_{21}+y_{21})^{n_{21}} \cdots (x_{2k}+y_{2k})^{n_{2k}} \cdots (x_{d \, k-1}+y_{d \, k-1})^{n_{d \, k-1}}(x_{dk}+y_{dk})^{n_{dk}} $$ $$=\sum_{m_{11}=0}^{n_{11}}\binom{n_{11}}{m_{11}}x_{11}^{m_{11}}y_{11}^{n_{11}-m_{11}} \cdots \sum_{m_{1k}=0}^{n_{1k}}\binom{n_{1k}}{m_{1k}}x_{1k}^{m_{1k}}y_{1k}^{n_{1k}-m_{1k}}\sum_{m_{21}=0}^{n_{21}}\binom{n_{21}}{m_{21}}x_{21}^{m_{21}}y_{21}^{n_{21}-m_{21}}\cdots \sum_{m_{2k}=0}^{n_{2k}}\binom{n_{2k}}{m_{2k}}x_{2k}^{m_{2k}}y_{2k}^{n_{2k}-m_{2k}}\cdots \sum_{m_{d \, k-1}=0}^{n_{d \, k-1}}\binom{n_{d \, k-1}}{m_{d \, k-1}}x_{d \, k-1}^{m_{d \, k-1}}y_{d \, k-1}^{n_{d \, k-1}-m_{d \, k-1}}\sum_{m_{dk}=0}^{n_{dk}}\binom{n_{dk}}{m_{dk}}x_{dk}^{m_{dk}}y_{dk}^{n_{dk}-m_{dk}}.$$
