Prove that the total number of arrangements of objects by taking any number of objects from $n$ different objects is $\lfloor e \times n! - 1 \rfloor$, where $e$ is the natural base.
I tried it by making cases that either we choose $1$ object in $\dbinom {n}{1} $ way and arrange it in $1$ way or two objects and so on, obtaining $$ \sum_{r=0}^n \dbinom {n}{r} \times r! $$
However, the quantity that we have to prove it equal to is entirely different.
Any help will be appreciated.
Thanks.
Hint: write the terms in your sum as $n!$ times something. Note that the sum of the somethings is close to the power series for $e$