If $$ f(n,m,k) = \begin{cases} n\cdot f(n-1,m,k+1) & \text{if } k = 0 \text{ or } k = 1 \\ (n+m)! & \text{if } k>m \\ n\cdot f(n-1,m,k+1) + m\cdot f(n,m-1,k-1) & \text{if } 1 < k \le m \end{cases} $$
Given that $$n > m$$ How can I prove that $$ \frac{f(n,m,0)}{(n+m)!} = \frac{n-m}{n+m} \text{ ?}$$