# sum of probability

I am having trouble of calculating the following probability:

Let $\epsilon_i$, $i=1,\dotsc,N$ be Rademacher random variables. Let $n_i\in \{0, 1, 2, \dotsc, 2M\}$, $i=1,\dotsc,N$ such that $\sum_{i=1}^N n_i=2M$.

Let $P_1=P\left(\left\{\prod_{i=1}^N\epsilon^{n_i}_i=1\right\}\bigcap\left\{\sum_{i=1}^N\epsilon_i=0\right\}\right)$. And $P_2=P\left(\left\{\prod_{i=1}^N\epsilon^{n_i}_i=-1\right\}\bigcap\left\{\sum_{i=1}^N \epsilon_i=0\right\}\right).$

I want to calculate, or bound from above, or approximate

$$\sum_{n_1+\cdots+n_n=2M, n_i\in\{0,1,\ldots,2M\}} P_1-P_2$$

Denoting by $A=\{i: n_i\text{ even}\}$, $B=\{i: n_i \text{ odd}\}$, $K=\operatorname{Card}(B)$. And observing that $\prod_{i=1}^N\epsilon^{n_i}_i=\prod_{i\in B}\epsilon^{n_i}_i=\prod_{i=1}^K\epsilon^{n_i}_i$, we can conclude that (see Probability of two events) $$P_1=2^{-N}\left(C_{N-K}^{N/2}+C_K^2C_{N-K}^{N/2-2}+\cdots+C_K^{N/2}\right)=2^{-N}C_1(N,K)$$ $$P_2=2^{-N}\left(C_{K}^{1}C_{N-K}^{N/2-1}+C_K^3C_{N-K}^{N/2-3}+\cdots+C_K^{K-1}C_{N-K}^1\right)=2^{-N}C_2(N,K)$$

I have problem now, how to calculate cardinality of $\operatorname{Card}(B)=K$ in terms of $n_i$. Would it be that $\operatorname{Card}(B)=K$ is how many ways to choose $n_i\in \{0,1,\ldots, 2M\}$, such that sum of them is odd?