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Counting Warmup

This analysis explores the composition of 7-bit binary strings that contain at least one run of three consecutive '1's or '0's. We denote the set of all 7-bit strings as S and focus on those subsets R that meet the criteria. By breaking down the strings into disjoint subsets based on the placement of the first block of '1's or '0's, we calculate the total number of valid strings using combinatorial principles. We find that there are 86 valid strings in total after removing overlaps between the sets containing runs of '1's and '0's.

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Counting Warmup

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  1. Counting Warmup

  2. Strings of 7 bits How many have a run of at least 3 1s? Let S={0,1}7 Let R⊆S be the strings with a run of 3 1s R={0,1}*111{0,1}*∩S To count R, split it into disjoint subsets and count the subsets Break R into subsets according to the position of the FIRST block of three 1s A string in R could be of the form 111xxxx, 0111xxx, x0111xx,xx0111x, or abc0111 where x is 0 or 1, and abc is any string of 3 bits except 111 16+8+8+8+7=47 Sum rule: If R=R1∪…∪Rn and the Ri are pairwise disjoint, then |R|=|R1|+…+|Rn|

  3. Strings of 7 bits How many members of S have a run of exactly 3 1s but not longer? Easiest to calculate how many have a run of at least four 1s and take the difference: 1111xxx, 01111xx, x01111x,xx01111 so20total 47-20=27 General rule: If A⊆B then |B-A|=|B|-|A|

  4. Strings of 7 bits How many members of S have a run of at least 3 1s or at least 3 0s? # with 3 1s + # with 3 0s - # with both (so they will not get counted twice) General rule: |A∪B|=|A|+|B|-|A∩B| Strings with both a run of 3 1s and 3 0s = 000111x, 111000x, 00001111, 1111000, 1000111, 0111000: 8 possibilities 47+47-8=86

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