Count number of binary strings

A string of binary digits of length $11$ bits with at least four $0$ 's and four $1$ 's has three bits leftover that could be either all three $0$ 's, two $0$ 's and one $1$ , two $1$ 's and one $0$ , or all three $1$ 's. These have the following total number of different anagrams:

$0000 \text{ } 1111 \text{ } 000 \rightarrow \frac{11!}{7!4!} = 330$

$0000 \text{ } 1111 \text{ } 001 \rightarrow \frac{11!}{6!5!} = 462$

$0000 \text{ } 1111 \text{ } 011 \rightarrow \frac{11!}{6!5!} = 462$

$0000 \text{ } 1111 \text{ } 111 \rightarrow \frac{11!}{7!4!} = 330$

for a total of $2(330 + 462) = \boxed{1584}$ different strings.