I want to find the number of permutations of the set $\{1,2,3,4,5,6\}$ if not one element is in that position as in the original input $1 2 3 4 5 6$.

Question

The answer given is $5 \times 4 \times 3 \times 2 \times 1 \times 1$.

I am trying to understand the solution. For the first position we must choose an element from the set $\{2,3,4,5,6\}$. Then for the next position we must choose from $\{1,3,4,5,6 \} - \{4\} = \{1,3,5,6\}$ assuming $4$ was chosen for the first position. etc.

I still do not feel comfortable with this. Maybe a different way to arrive at the same solution might be useful.

Answer 1

This is just an application of derangement, in other words the permutation can have no fixed points. The formula for derangement of $n$ elements is: $!n=n!(\frac{1}{2!}-\frac{1}{3!}+\ldots+(-1)^{n}\frac{1}{n!})$.