(a) How many bit strings of length 8 contain at least 6 ones? (b) How many bit strings of length 8 contain at least 3 ones and 3 zeros?

a) We need to consider 3 different cases.

Case 1: there are six 1s.

$C(8, 6)=\dbinom{8}{6}$

Case 2: there are seven 1s.

$C(7, 6)=\dbinom{8}{7}$

Case 3: there are eight 1s.

$C(8, 8)=\dbinom{8}{8}$

Since there is no overlap between these three cases, the addition principle is used to determine the total number of 8-bit strings that contain at least six 1s: 

$C(8, 6)+C(8,7)+C(8,8)=\dbinom{8}{6}+\dbinom{8}{7}+\dbinom{8}{8}$

$=28+8+1=37$

37 bit strings of length 8 contain at least 6 ones.

b)  The amount of strings which contain at least 3 ones and 3 zeros is the sum of the amounts of the strings that contain exactly 3, 4, 5 ones (other places will be automatically taken by zeros)

$C(8, 3)+C(8,4)+C(8,5)=\dbinom{8}{3}+\dbinom{8}{4}+\dbinom{8}{5}$

$=56+70+56=182$

182 bit strings of length 8 contain at least 3 ones and 3 zeros.