Ok I have a basic exercise that wants me to explain in my own words the following :
$a) \ \ L_{k}^{or}(N)=N\cdot L_{k-1}^{or}(N) $
$b) \ \ L_{k}^{os}(N)=(N-k+1)\cdot L_{k-1}^{os}(N) $
$c) \ \ L_{k}^{ns}(N) \cdot L_{k}^{os}(k)= L_{k}^{os}(N)$
$\cdot \ \textbf{or}$ means $\textbf{ordered}$ samples $\textbf{with}$ replacement
$\cdot \ \textbf{os}$ means $\textbf{ordered}$ samples $\textbf{without}$ replacement
$\cdot \ \textbf{nr}$ means $\textbf{non-ordered}$ samples $\textbf{with}$ replacement
$\cdot \ \textbf{ns}$ means $\textbf{non-ordered}$ samples $\textbf{without}$ replacement
$-$ The first one is pretty easy, if I pickup $k$ samples from $N$ and replace it every-time, I still have $N$ choices next time.
$-$ The second one too, if I pickup $k$ samples from $N$ and I don't replace them in the bag, I will have $N-(k-1)$ choices the next time $\iff$ $N-k+1$
$-$ But I have no idea for the third one. I struggle a lot with intuitive thinking in combinatorics, probability etc.. Any help would be great !
I assume all objects as distinct. In third case, we are selecting without replacement.
LHS means that you choose $k$ objects from $N$ objects without considering order, and after choosing you order them.
RHS means that you choose $k$ objects from $N$ objects, but you also consider the order of these $k$ objects while selection.
Alternatively
We have that $L_{k}^{ns}(N) = \binom{N}{k}$ and $L_{k}^{os}(k) = k!$. Also $L_{k}^{os}(N) = k! \binom{n}{k}$