What shown below is a reference from Introduction to vectors and tensors by Ray M. Bowen and C.C Wang. Precisely it is de definition of the generalised Kroneker delta thus if you know it you can even not read what is written in the image and so you can directly read what I ask to prove.
So using the sequence without repetition it is possible to prove that $$ \sum_{i_1,i_2,...,i_K=1}^N\delta^{i_1,i_2,...,_K}_{i_1,i_2,...,i_K}=\frac{N!}{(N-K)!} $$ thus my text asks to prove as exercise that $$ \sum_{i_{R+1},i_{R+2},...,i_K=1}^N\delta^{i_1,...,i_R,i_{R+1},...,i_K}_{j_1,...,j_R,i_{R+1},...,i_K}=\frac{(N-R)!}{(N-K)!}\delta^{i_1,...,i_R}_{j_1,...,j_R} $$ but unfortunately using the preceding formula with $(N-R)$ instead of $N$ we have $$ \sum_{i_{R+1},i_{R+2},...,i_K=1}^N\delta^{i_1,...,i_R,i_{R+1},...,i_K}_{j_1,...,j_R,i_{R+1},...,i_K}=\frac{(N-R)!}{((N-R)-K)!}\delta^{i_1,...,i_R}_{j_1,...,j_R} $$ that is different from what indicated above. So could someone help me, please?
Your $N\mapsto N-R$ is invalid because, for its right-hand side to be right, you'd also need to cap the indices summed over on the left-hand side at $N-R$. The correct reasoning is that each of the $\binom{N-R}{K-R}$ choices of $i_{R+1},\,\cdots i_K$ consistent with a given choice of the $i_1,\,\cdots,\,i_R$ admits $(K-R)!$ permutations, contributing to the sum iff $\delta^{i_1\cdots i_R}_{j_1\cdots j_R}=1$. So it evaluates to$$\frac{(N-R)!}{(K-R)!(N-K)!}(K-R)!\delta_{j_1\cdots j_R}^{i_1\cdots i_R}=\frac{(N-R)!}{(N-K)!}\delta_{j_1\cdots j_R}^{i_1\cdots i_R}.$$