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Counting Warmup

Counting Warmup. 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

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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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