How many bit strings of length 11 have more 0s than 1s?

The order of bits is not important since we are interested in the number of ones, not the order of ones. Thus, we need to use combinations.

$\displaystyle C(n,r) = \frac{n!}{r! (n-r)!}$

$n=11$. There are more zeros than ones when less than 6 bits have ones.

$\displaystyle C(11,5) = \frac{11!}{6!\,5!} =462$

$\displaystyle C(11,4) = \frac{11!}{7!\,4!} =330$

$\displaystyle C(11,3) = \frac{11!}{9!\,3!} =165$

$\displaystyle C(11,2) = \frac{11!}{9!\,2!} =55$

$\displaystyle C(11,1) = \frac{11!}{10!\,1!} =11$

$C(11,0)=1$

So there are $462+330+165+55+11+1=1024$ bit strings of length 11 that have more 0s than 1s.

Answer: 1024