According to OEIS, the formula for getting the number of arrangements of any subset of n distinct objects is:
$$\sum_{k=0}^n{n \choose k}k! = \sum_{k=0}^n\frac{n!}{k!(n-k)!}k!\\ = n!\sum_{k=0}^n \frac{1}{(n-k)!} = n! \sum_{k=0}^n \frac{1}{k!}$$
From the first expression to the second expression is the formula for n choose k. From the second expression to the third expression, k! is cancelled out. But next why do we have $\frac{1}{(n-k)!} = \frac{1}{k!}$?
Let's see what happens for Sigma $ n!\sum_{k=0}^n \frac{1}{(n-k)!} = n! \sum_{k=0}^n \frac{1}{k!} $ its suffice to show $$\sum_{k=0}^n \frac{1}{(n-k)!} = \sum_{k=0}^n \frac{1}{k!}$$ so put the numbers $$\sum_{k=0}^n \frac{1}{(n-k)!}=\frac 1{(n-0)!}+\frac 1{(n-1)!}+\frac 1{(n-1)!}+\cdots+\frac 1{(n-n)!}=\\\frac 1{(n)!}+\frac 1{(n-1)!}+\frac 1{(n-1)!}+\cdots+\frac 1{(0)!}$$ and other Sigma $$\sum_{k=0}^n \frac{1}{k!}=\frac 1{0!}+\frac 1{1!}+\frac 1{2!}+\cdots+\frac 1{(n-1)!}+\frac 1{n!}$$ and both are the same