The famed matching problem, up a notch

If there was an easier way, I'd be happy to hear about it.

Given that we have $7$ labeled envelopes, we are to derange $7$ letters, and with respect to both the letters and the envelopes, we have to derange $7$ pictures as well.

It is quite intuitive to think that we can first carry out the derangements of letters with respect to the envelopes first, which is $!7 = 7! - {7 \choose1}6! + {7 \choose 2}5! - ... = 1854$

Now, let us look at its behavior when we only have $4$ of each.

If we have $4$ each of envelopes, letters, and pictures, we'll have $9$ derangements of letters to envelopes, and $9$ derangements of pictures to envelopes as well. How do we make sure now that no envelope contains a letter and its corresponding picture? We now plot them against each other; that is, we check all the $9 \times 9 = 81$ permutations and check which of these do not share a common label for a picture and a letter at any position (We now know that all of these arrangements are deranged with respect to the envelopes).

To put that into a clearer perspective, we check any two of the nine derangements (order considered, of course) whether any of the envelopes contain a correct pair of letter and picture. For example, for envelopes sequenced $\{A,B,C,D\}$ , the letters may be placed like this: $\{B,D,A,C\}$ . The pictures may also be placed like $\{D,C,A,B\}$ , but since envelope $C$ will contain both letter $A$ and picture $A$ , this doesn't count as a valid way. All in all, there are $24$ of them, out of 81.

And that's what I did for $1854 \times 1854$ permutations. And got me $\boxed{1073760}$ .

tl;dr Compare each derangement with all the other derangements and count which pairs do not contain any same letter at any position.