Conditions for equality of two binomial sums

Question

Let $k,r,n$ be integers such that $0<k,r<n$. Let

$$K=\sum^n_{i=k}k^{n-i}\binom{n-k}{i-k}^2k!(i-k)! \,\text{ and }\, R=\sum^n_{i=r}r^{n-i}\binom{n-r}{i-r}^2r!(i-r)!.$$

How to show that "$K=R\:\Rightarrow\: k=r$"? But if it does not holds, I want to see a counter example.

Answer 1

For $(k, r, n) = (1, 2, 2)$, we have \begin{align*} K = \sum_{i=1}^2 \binom{1}{i-1} (i-1)! = \binom{1}{0} + \binom{1}{1} = 2 \end{align*} and \begin{align*} R = \sum_{i=2}^22^{2-i} \; 2!\;(i-2)! = 2. \end{align*}