A contingency table is an $m\times n$ matrix of nonnegative integers $\left(a_{\mathrm{ij}}\right)$ with given row sums $R_i\=\sum _{j\=1}^n{a}_{\mathrm{ij}}$ and column sums $C_j\=\sum _{i\=1}^m{a}_{\mathrm{ij}}$. Given the $m$-vector $R$ and $n$-vector $C$, we want to compute the number of distinct contingency tables. Use a bounded-composition with $M=C$ and $T=R_1$ to generate a valid first row. Use a bounded-composition with $M\=C-\sum _{k\=1}^{i-1}{r}_k$ and $T=R_i$, where $r_k$ is the content of the $k$-th row, to generate a valid $i$-th row. The last row is fixed by preceding rows. The following procedure uses a recursive procedure, cnt, to generate and enumerate the iterators for each row.

Cnt := proc(R :: ~Array, C :: ~Array)
local cnt,m;
    if add(R) <> add(C) then
        error "row and column sums must be equal";
    end if;
    m := numelems(R);
    cnt := proc(i, c)
    local r;
        if i = m then
            return 1;  # final row
        else
            add(cnt(i+1, c-r), r = BoundedComposition(c,R[i]));
        end if;
    end proc;
    cnt(1, C);
end proc:

Try it with a known case (value must be 5!).

$\mathrm{Cnt}\left(\mathrm{`\$`}\left(\left[\mathrm{`\$`}\left(1\&comma;5\right)\right]\&comma;2\right)\right)$

$120$

Try a simple case

$\mathrm{Cnt}\left(\left[2\&comma;3\&comma;4\right]\&comma;\left[1\&comma;3\&comma;5\right]\right)$

$24$