Problem 3: (15 points) Consider a binary code with 8 bits (O or 1) in each code

Question

Problem 3: (15 points) Consider a binary code with 8 bits (O or 1) in each code word. An example of a code word is 01010101. (1) If the order of bits is important, how many different code words are there? (2) If the order of bits is important, how many code words have exactly 4 zeros? (3) Suppose that only the number of ones and zeros in the code word is of importance. In other words, the code word 01010101 is treated the same as 10101010, 1110000, etc.. How many code words are there in this case?

Explanation / Answer

since there is 8 bits code and each place of the code is placed by two way i.e. with 0 or 1 then

(a)answer is 256

first place of code is filled by either 0 or 1 i.e. 2 way, so second , third , .....,8th place of code filled by 2 way so

required number of code=2*2*2*2*2*2*2*2=28=256

(b)answer is 70

there are 8 bits , and it is filled by exactly 4 zeros, it means remaining 4 bits is filled by other bit

4 places of 8 bits code can be filled by 0 with 8C4 =70ways

remaining 4 places of 8 bits code can be filled by 1 with 4C4=1 way

so required number of code words=8C4 x 4C4=70*1=70

(c) answer is 9

there will be 9 code word as

8 zerors, 1zeros+7 ones, 2zeros+6 ones,3zeros+5 ones,4zeros+4 ones, 5zeros+3 ones, , 6zeros+2 ones, 7 zeros+ 1 one, 8 ones

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