Answer to Question #186623 in Discrete Mathematics for Miucce

Question

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

Answer 1

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