11. A farmer has 100 pigs and 100 chickens. He has four adjacent square yards, forming a $2 \times 2$ grid. The farmer wants to distribute the livestock among the yards according to the following requirements: the first row has 120 heads, the second row has 300 feet; the first column has 100 heads, the second column has 320 feet. Then there are $\qquad$ different ways to distribute them.
11.341.

As shown in Figure 8, label the four courtyards with letters $A_{i} (i=1,2,3,4)$. Let $A_{i}$ courtyard have $x_{i}$ pigs and $y_{i}$ chickens.
From the problem, we get
every 8
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}=100, \\
y_{1}+y_{2}+y_{3}+y_{4}=100, \\
x_{1}+y_{1}+x_{2}+y_{2}=120, \\
4\left(x_{3}+x_{4}\right)+2\left(y_{3}+y_{4}\right)=300, \\
x_{1}+y_{1}+x_{3}+y_{3}=100, \\
4\left(x_{2}+x_{4}\right)+2\left(y_{2}+y_{4}\right)=320 .
\end{array}\right.
$$

Solving, we get $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=(x, 30-x, 40-x, 30+x)$,
$$
\left(y_{1}, y_{2}, y_{3}, y_{4}\right)=(y, 90-y, 60-y, y-50) \text {, }
$$

where $x$ has 31 possible values from $0$ to $30$, and $y$ has 11 possible values from $50$ to $60$.

Therefore, there are $11 \times 31=341$ ways to distribute the livestock to the courtyards as required.