# Finding the number of permutations of the digits 1 through 9 in which none of the blocks 34, 45 and 738 appears

How do I find the number of permutations of the digits 1 through 9 in which none of the blocks 34, 45 and 738 appears?

I know that I can probably brute force the solution, but is there a concept I can apply here that makes finding the solution quicker?

ATTEMPT AT SOLUTION:

The total number of permutations of digits 1 through 9 is $9!$

The block 34 can occur in $8!$ ways if we treat "34" as a single symbol in an 8 digit alphabet. Similarly, the block 45 can occur $8!$ ways and the block 738 can occur $7!$ ways.

*The blocks 34 and 45 can both occur together if the block 345 is in the permutation. This can occur $7!$ times.
The blocks 34 and 738 cannot both occur together since 3 can only be used once.
The blocks 45 and 738 can occur together $6!$ ways by treating 34 and 738 as single symbols in a 6 digit alphabet.

The blocks 34, 45, and 738 cannot all occur together.

Therefore, by Inclusion-Exclusion, the total number of permutations satisfying the conditions is given by: $$9! - (8! + 8! + 7!) + (7! + 0 + 6!) - (0) = 282,960$$

Does my logic look alright? Particularly the line denoted by the asterisk *.