(11) (20 points) Form a seven-digit number using the digits $1, 2, 3, 4, 5, 6, 7$, such that no three of the four odd numbers are adjacent. How many such seven-digit numbers are there?
(11) Subtract the number of the following two types of seven-digit numbers from all possible seven-digit numbers $\mathrm{A}_{7}^{7}$.
(1) Seven-digit numbers where four odd numbers are adjacent. First, consider the four odd numbers as a single group and arrange them with the remaining three even numbers, which can be done in $\mathrm{A}_{4}^{4}$ ways; then arrange these four odd numbers, which can be done in $\mathrm{A}_{4}^{4}$ ways. There are $\mathrm{A}_{4}^{4} \cdot \mathrm{A}_{4}^{4}$ such seven-digit numbers.
(2) Seven-digit numbers where only three odd numbers are adjacent. First, arrange the even numbers, which can be done in $\mathrm{A}_{3}^{3}$ ways; then divide the four odd numbers into two groups, one with three and one with one, which can be done in $\mathrm{C}_{4}^{1}$ ways; then insert these two groups of odd numbers into the four gaps between the even numbers, which can be done in $\mathrm{A}_{4}^{2}$ ways. There are $\mathrm{A}_{3}^{3} \cdot \mathrm{C}_{4}^{1} \cdot \mathrm{A}_{4}^{2} \cdot \mathrm{A}_{3}^{3}$ such seven-digit numbers.
In summary, the number of seven-digit numbers that meet the conditions is
$$
\begin{aligned}
& \mathrm{A}_{7}^{7}-\mathrm{A}_{4}^{4} \cdot \mathrm{A}_{4}^{4}-\mathrm{A}_{3}^{3} \cdot \mathrm{C}_{4}^{1} \cdot \mathrm{A}_{4}^{2} \cdot \mathrm{A}_{3}^{3} \\
= & 5040-(576+1728) \\
= & 2736(\text { numbers). }
\end{aligned}
$$