Answer to Question #254456 in Discrete Mathematics for Fiez

Question #254456
(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?
1
Expert's answer
2021-10-21T13:27:18-0400

a) We need to consider 3 different cases.

Case 1: there are six 1s.


C(8,6)=(86)C(8, 6)=\dbinom{8}{6}

Case 2: there are seven 1s.


C(7,6)=(87)C(7, 6)=\dbinom{8}{7}

Case 3: there are eight 1s.


C(8,8)=(88)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)=(86)+(87)+(88)C(8, 6)+C(8,7)+C(8,8)=\dbinom{8}{6}+\dbinom{8}{7}+\dbinom{8}{8}

=28+8+1=37=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)=(83)+(84)+(85)C(8, 3)+C(8,4)+C(8,5)=\dbinom{8}{3}+\dbinom{8}{4}+\dbinom{8}{5}

=56+70+56=182=56+70+56=182

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



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