首先$n$个炮在一行操作一次的方案数为$2(n-2)$:前面两个炮只能向右吃，最后两个跑只能向左吃，而其余的炮既可以向左也可以向右，于是有$4+2(n-4)$种

于是操作$m$次的操作排列的方案数为$2^m(n-2)(n-3)...(n-m-1)=2^m\frac{(n-2)!}{(n-2-m)!}$

设$n=a-2,m=b-2$

首先选择从$k$选择$i$次操作第一行，选择$k-i$次操作第二行，而问题一还需要从$k$个位置选择出$i$个按序放置操作序列，而问题二必须强制第一行操作放置在第二行操作前面，由此有：

对于问题1即是求
∑0≤i≤k(ki){2in!(n−i)!}×{2k−im![m−(k−i)]!}=2k∑0≤i≤kk!i!(k−i)!⋅n!(n−i)!⋅m![m−(k−i)]!=2kk!∑0≤i≤k(ni)(mk−i)=2kk!(n+mk)\begin{aligned} &\sum_{0\leq i\leq k}\dbinom{k}{i}\{2^i\frac{n!}{(n-i)!}\}×\{2^{k-i}\frac{m!}{[m-(k-i)]!}\}\\ \\&=2^k\sum_{0\leq i\leq k}\frac{k!}{i!(k-i)!}·\frac{n!}{(n-i)!}·\frac{m!}{[m-(k-i)]!}\\ \\&=2^kk!\sum_{0\leq i\leq k} \dbinom{n}{i}\dbinom{m}{k-i}\\\\&=2^kk!\dbinom{n+m}{k} \end{aligned}​0≤i≤k∑​(ik​){2i(n−i)!n!​}×{2k−i[m−(k−i)]!m!​}=2k0≤i≤k∑​i!(k−i)!k!​⋅(n−i)!n!​⋅[m−(k−i)]!m!​=2kk!0≤i≤k∑​(in​)(k−im​)=2kk!(kn+m​)​

对于问题2即是求
∑0≤i≤k{2in!(n−i)!}×{2k−im![m−(k−i)]!}=2k∑0≤i≤kn!(n−i)!×m![m−(k−i)]!=2kn!m!(n+m−k)!∑0≤i≤k(n+m−k)!(n−i)![m−(k−i)]!=2kn!m!(n+m−k)!∑0≤i≤k(n+m−kn−i)=2kn!m!(n+m−k)!∑n−k≤i≤n(n+m−kj)\begin{aligned} &\sum_{0\leq i\leq k}\{2^i\frac{n!}{(n-i)!}\}×\{2^{k-i}\frac{m!}{[m-(k-i)]!}\}\\\\&=2^k\sum_{0\leq i\leq k}\frac{n!}{(n-i)!}×\frac{m!}{[m-(k-i)]!}\\\\&=2^k\frac{n!m!}{(n+m-k)!}\sum_{0\leq i\leq k}\frac{(n+m-k)!}{(n-i)![m-(k-i)]!}\\\\&=2^k\frac{n!m!}{(n+m-k)!}\sum_{0\leq i\leq k}\dbinom{n+m-k}{n-i}\\\\&=2^k\frac{n!m!}{(n+m-k)!}\sum_{n-k\leq i\leq n}\dbinom{n+m-k}{j} \end{aligned}​0≤i≤k∑​{2i(n−i)!n!​}×{2k−i[m−(k−i)]!m!​}=2k0≤i≤k∑​(n−i)!n!​×[m−(k−i)]!m!​=2k(n+m−k)!n!m!​0≤i≤k∑​(n−i)![m−(k−i)]!(n+m−k)!​=2k(n+m−k)!n!m!​0≤i≤k∑​(n−in+m−k​)=2k(n+m−k)!n!m!​n−k≤i≤n∑​(jn+m−k​)​
其中下面一项是组合数种的连续多项之和，考虑前缀和！

∑n−k≤j≤n(n+m−kj)=∑0≤j≤n(n+m−kj)−∑0≤j≤n−k−1(n+m−kj)=∑0≤j≤n(n+m−kj)−∑0≤j≤n−k−1(n+m−kn+m−k−j)=∑0≤j≤n(n+m−kj)−∑m+1≤j≤n+m−k(n+m−kj)=∑0≤j≤n(n+m−kj)−{∑0≤j≤n+m−k(n+m−kj)−∑0≤j≤m(n+m−kj)}=∑0≤j≤n(n+m−kj)+∑0≤j≤m(n+m−kj)−2n+m−k=sum(n+m−k)\begin{aligned} &\sum_{n-k\leq j\leq n}\dbinom{n+m-k}{j}\\\\&=\sum_{0\leq j\leq n}\dbinom{n+m-k}{j}-\sum_{0\leq j\leq n-k-1}\dbinom{n+m-k}{j}\\\\&=\sum_{0\leq j\leq n}\dbinom{n+m-k}{j}-\sum_{0\leq j\leq n-k-1}\dbinom{n+m-k}{n+m-k-j}\\\\&=\sum_{0\leq j\leq n}\dbinom{n+m-k}{j}-\sum_{m+1\leq j\leq n+m-k}\dbinom{n+m-k}{j}\\\\&=\sum_{0\leq j\leq n}\dbinom{n+m-k}{j}-\{\sum_{0\leq j\leq n+m-k}\dbinom{n+m-k}{j}-\sum_{0\leq j\leq m}\dbinom{n+m-k}{j}\}\\\\&=\sum_{0\leq j\leq n}\dbinom{n+m-k}{j}+\sum_{0\leq j\leq m}\dbinom{n+m-k}{j}-2^{n+m-k}=\text{sum}(n+m-k) \end{aligned}​n−k≤j≤n∑​(jn+m−k​)=0≤j≤n∑​(jn+m−k​)−0≤j≤n−k−1∑​(jn+m−k​)=0≤j≤n∑​(jn+m−k​)−0≤j≤n−k−1∑​(n+m−k−jn+m−k​)=0≤j≤n∑​(jn+m−k​)−m+1≤j≤n+m−k∑​(jn+m−k​)=0≤j≤n∑​(jn+m−k​)−{0≤j≤n+m−k∑​(jn+m−k​)−0≤j≤m∑​(jn+m−k​)}=0≤j≤n∑​(jn+m−k​)+0≤j≤m∑​(jn+m−k​)−2n+m−k=sum(n+m−k)​

考虑求出$S_{n,m}={\sum }_{i=0}^m\left(\genfrac{}{}{0ex}{}{n}{i}\right)$的关于$n$的递推式

$$\begin{array}{rl}{S}_{n,m}=\sum _{i=0}^m\left(\genfrac{}{}{0ex}{}{n}{i}\right)& =\sum _{i=0}^m[(\genfrac{}{}{0ex}{}{n-1}{i})+(\genfrac{}{}{0ex}{}{n-1}{i-1})]\\ & =S_{n-1,m}+S_{n-1,m-1}\end{array}$$

而$$\begin{array}{rl}{S}_{n-1,m}=\sum _{i=0}^m\left(\genfrac{}{}{0ex}{}{n-1}{i}\right)& =S_{n-1,m-1}+\left(\genfrac{}{}{0ex}{}{n-1}{m}\right)\end{array}$$

带入可得出
$$S_{n,m}=2S_{n-1,m}-\left(\genfrac{}{}{0ex}{}{n-1}{m}\right)$$

由此可得出
$$\begin{array}{rl}\text{sum}(x)& =\sum _{0\le j\le n}\left(\genfrac{}{}{0ex}{}{x}{j}\right)+\sum _{0\le j\le m}\left(\genfrac{}{}{0ex}{}{x}{j}\right)-2^x\\ & \\ & =S_{x,n}+S_{x,m}-2^x\\ & \\ & =2\text{sum}(x-1)-\left(\genfrac{}{}{0ex}{}{x-1}{n}\right)-\left(\genfrac{}{}{0ex}{}{x-1}{m}\right)\end{array}$$