Derangement of Odd Objects

Question

I was recently learning the concept of derangement of $n$ objects and I did not want to mug up the formula so instead, I solved by inclusion-exclusion principle.
The question was to find the number of ways in which $4$ envelopes none of them can be placed in the correct place. I know it can be solved by the derangement formula $$n! \left(1- \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}+ (-1)^n\frac{1}{n!} \right ).$$ and for $n=4$ we get, $$n(D) = 9$$ I solved this by the concept of inclusion-exclusion and long story short, I got this formula, $$n(D) = n! - \sum ^ {n} _{k=1} \binom{n}{k}.$$ However I am not able to use this when $n$ is of the form $2n-1$ $\forall \; n \gt 2$. This I realized is because we cannot derange $1$ object only out of $3,5,7,9...2n-1$ objects.
So can anyone tell me how to go about how to derive a formula like this for $2n-1$ objects.

Answer 1

I got the formula $n!-\sum_{k=1}^n \binom{n}k.$

This is incorrect. The correct formula is $n!-\sum_{k=1}^n (-1)^{k+1}\binom{n}k (n-k)!$.

we cannot derange $1$ object out of $3,5,\dots,2n-1$.

This has nothing to do with the problem at hand. Sure, it is impossible to choose a permutation so exactly one object is out of place. This is also true when the number of objects is even. But the problem is about deranging all of the objects, not just one of them, so this does not matter.