Show that the following two inequalities have the same number of integer solutions.
(A) $|x_1|+|x_2|+...+|x_n| \le m$
(B) $|y_1|+|y_2|+...+|y_m| \le n$, where m and n are two positive integers.
Well, I tried to solve this by using lattice paths argument which is effective to show that A and B have the same number of non-negative solutions but failed. I want to know bijective proofs and also closed form of the number of solutions if it exists.
Consider the inequality $$x_1+x_2+\dots+x_n \le m,$$
where $x_1,\dots,x_n$ are integer variables. We know that the numbers of positive and non-negative solution vectors $(x_1,\dots,x_n)$ satisfying this inequality are $\binom{m}{n}$ and $\binom{m+n}{n}$, respectively (see e.g., LINK-1 and LINK-2).
The number of non-negative integer solutions to $$|x_1|+|x_2|+\dots+|x_n| \le m,$$ is the same as the number of non-negative integer solutions to $$x_1+x_2+\dots+x_n \le m,$$
which is equal to $\binom{m+n}{n}$. For the same reason, the number of non-negative integer solutions to $$|x_1|+|x_2|+\dots+|x_m| \le n,$$ is equal to $\binom{m+n}{m}$. Since we have
$$\binom{m+n}{m} = \binom{m+n}{n},$$
we can conclude that the numbers of non-negative solutions to $\sum_{i=1}^{n} |x_i| \le m$ and $\sum_{i=1}^{m} |x_i| \le n$ are the same.
The general case:
Consider a non-negative solution of $\sum_{i=1}^{n} |x_i| \le m$ in which $1 \le k \le \min(m,n)$ variables are positive ($k>m$ is impossible). The number of such solutions is equal to the number of ways we can select $k$ variables from $n$ variables times the number of positive solutions to
$$x_1+x_2+\dots+x_k \le m.$$
Since there are $\binom{n}{k}$ ways to select $k$ variables from $n$ variables and there are $\binom{m}{k}$ positive solutions to the last inequality, there are $\binom{n}{k}\binom{m}{k}$ non-negative solutions to $\sum_{i=1}^{n} |x_i| \le m$ with $k$ positive variables. On the other hand, there are $2^k$ number of solutions corresponding to each of these solutions that can be achieved by changing sign of the variables. So the total number of integer solutions to $\sum_{i=1}^{n} |x_i| \le m$ would be
$$N_1 = 1+\sum_{k=1}^{\min(m,n)} 2^k \binom{n}{k}\binom{m}{k}.$$
Note that we added $1$ as we also need to consider the solution vector $(x_1,\dots,x_n) = (0,\dots,0)$. For the same reason, the number of solutions to $\sum_{i=1}^{m} |x_i| \le n$ is equal to
$$N_2 = 1+\sum_{k=1}^{\min(m,n)} 2^k \binom{m}{k}\binom{n}{k}.$$ It is clear that $N_1 = N_2$.
The following generating function approach proves the symmetry, but does not give a final formula as simple as Opt's answer.
We begin with $$x_1+x_2+\ldots+x_n\leq m\ .\tag{1}$$ Consider the set of all functions $$f:\quad [n]\to{\mathbb Z}, \qquad k\mapsto x_k\ .\tag{2}$$ The value $v(f)$ of such a function is defined to be $$v(f):=z^{|x_1|+\ldots+|x_n|}\ ,$$ where $z$ is an indeterminate. If $n=1$ the total value of all $f:\>[1]\to{\mathbb Z}$ is given by the formal power series $$1+2z+2z^2+2z^3+\ldots={1+z\over1-z}\ .$$ The distributive law then implies that for arbitrary $n\geq1$ the total value of all $f$ in $(2)$ is given by $$S:=\left({1+z\over1-z}\right)^n\ .$$ Each $f$ with $\sum_{k=1}^n |x_k|=m$ contributes $z^m$ to $S$, hence the number of these $f$s is the coefficient of $z^m$ in $S$. Since we want the number of solutions of the inequality $(1)$ we multiply $S$ by $1+z+z^2+\ldots={1\over1-z}$ and now determine the coefficient of $z^m$ in $$S':={(1+z)^n\over(1-z)^{n+1}}=\sum_{j=0}^n{n\choose j}z^j\cdot\sum_{k=0}^\infty{n+k\choose k} z^k\ .$$ This coefficient comes to $$\sum_{j+k=m}{n\choose j}{n+k\choose k}=\sum_{j=0}^{\min\{m,n\}}{n\choose j}{n+m-j\choose m-j}\ .$$ Chasing factorials allows to rewrite the last sum as $${m+n\choose m}\sum_{j=0}^{\min\{m,n\}}{{m\choose j}{n\choose j}\over{m+n\choose j}}\ ,$$ which is obviously symmetric in $m$ and $n$.
