Problem 11. (2 points) How many 12-bit strings (that is, bit...

Problem 11. (2 points) How many 12-bit strings (that is, bit strings of length 12) are there which: a. Start with the sub-string 011? b. Have weight 4 (i.e., contain exactly 4 1's) and start with the sub-string 011? c. Either start with 011 or end with 01 (or both)? d. Have weight 4 and either start with 011 or end with 01 (or both)? Note: You can earn partial credit on this problem.

Transcript

00:01 Hello students we need to find how many 12 -bate strings are there in which first one given starts with sub string 011 that is for the first three positions are filled with 0 1 1 and the remaining 9 positions are empty after three positions occupied we still have nine remaining positions to be filled and these positions can be filled either by one or a zero therefore each blank has two ways and we have nine positions therefore we have two rise to the power nine ways are available therefore we have two rise to the power nine is equal to five and two ways ways can be filled with the rest three rest nine positions next now we have the weight for that contains exactly four ones and start with the substring zero one given according to the question first three positions are filled with zero one one and the nine positions are empty and weight is four which means the whole string should contain four one here already two ones are there therefore we have to fill the remaining positions with two ones that is 4 minus 2 is equal to 2 once let us now assume that out of 9 positions 2 positions are filled with 1s to satisfy the weight then we have 9 minus 2 is equal to 7 positions and these 7 position should be filled with 0 and this can be done by 9 2 ways which gives the value 9 factorial by 2 factorial into 9 minus 2 factorial that is 9 by 2 permutations value on simplifying this we get that is we have 36 ways next either start with 011 or end with 0 1 or both now let us assume that start with 011 b a and or end with 0 1 b.

03:01 Therefore we need according to the union formula we have n of a union b is equal to n of a plus n of b minus n of a intersection b using the union formula.

03:19 Now we have n of a is equal to starting with 011 we have 9 ways and we can be filled with 0 or 1 therefore 2 rise to the power 9 and n of b we have 2 rise to the power 10 since last 2 positions are filled and we are left out with 10 position with the 2 ways of filling now n of a intersection b is equal to 2 rise to the power 7 since starting 011 and the ending 0 1 5 positions are filled and and left with 7 position with the two ways of filling.

03:59 That is therefore now substituting in the formula, n of a union b is equal to 2 rise to the power 9 plus 2 rise to the par 10 minus 2 rise to the power 7.

04:09 This implies we have 2 rise to the part 7 into 4 plus 8 minus 1.

04:15 On simplifying we have therefore 1408 ways.

04:24 Next, given half weight 4...

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