Number of bit string interpretations correct?

Question

Suppose you are given a bit string $B[1 ... n]$. Now, suppose that some bits are just padding bits conveying no information, the rest of the bits may be permuted and some meaningful (that is, non-padding) bits should be inverted from $0$ to $1$ and vice versa.

According to my intuition, the total number of interpretations is $$ I(B) = \sum_{i = 0}^n \binom{n}{i} i! 2^i. $$ The idea is that we choose all bit combinations, one by one. Then, we permute each combination. Finally, we multiply by $2^i$ in order to take the bit inversions into account.

Is this formula correct?

Example

$p$ as padding, meaningless bit. $b$ as non-padding, meaningful bit. $\bar{b}$ as meaningful, inverted bit.

1. $pp$
2. $pb$
3. $p\bar{b}$
4. $bp$
5. $\bar{b}p$
6. $bb^\prime$
7. $b\bar{b^\prime}$
8. $\bar{b}b^\prime$
9. $\bar{b}\bar{b^\prime}$
10. $b^\prime b$
11. $b^\prime \bar{b}$
12. $\bar{b^\prime}b$
13. $\bar{b^\prime}\bar{b}$

Note that $I(B[1 … 2]) = 13$.

Answer 1

When there are $i$ significant bits, you can form $2^i$ arrangements that can be permuted in $i!$ ways, and interspersed each with the remaining $n-i$ bits. Hence in total

$$\sum_{i=0}^n2^ii!\binom ni.$$

E.g., for $n=2$,

$$1\cdot1\cdot1+2\cdot1\cdot2+4\cdot2\cdot1=13.$$

This just confirms your findings.