7. The seven-digit numbers formed by the 7 digits $1,2,3,4,5,6,7$ without repetition, where the three even numbers are adjacent, and the four odd numbers are not all adjacent, have $\qquad$ numbers.
7. 432 Bind three even numbers together as one element $A$.

First, make a full permutation of the four odd numbers, then insert $A$ into the gaps between these four odd numbers, and finally make a full permutation of the elements in $A$. Therefore, there are $4! \times 3 \times 3! = 432$ seven-digit numbers that meet the problem's requirements.