let A={1xxxxxxx}where x∈{0,1}\text {let }A=\{1xxxxxxx\}\text{where }x\isin\{0,1\}let A={1xxxxxxx}where x∈{0,1}
∣A∣=27=128\vert{A}\vert=2^7=128∣A∣=27=128
let B={xxxxxx00}where x∈{0,1}\text {let }B=\{xxxxxx00\}\text{where }x\isin\{0,1\}let B={xxxxxx00}where x∈{0,1}
∣B∣=26=64\vert{B}\vert=2^6=64∣B∣=26=64
A∩B={1xxxxx00}where x∈{0,1}A\cap{}B=\{1xxxxx00\}\text{where }x\isin\{0,1\}A∩B={1xxxxx00}where x∈{0,1}
∣A∩B∣=25=32\vert{A\cap{}B}\vert=2^5=32∣A∩B∣=25=32
according to Inclusion-exclusion principle ∣A∪B∣=∣A∣+∣B∣−∣A∩B∣\text{according to Inclusion-exclusion principle }\vert{A\cup{}B}\vert=\vert{A}\vert+\vert{B}\vert-\vert{A\cap{}B}\vertaccording to Inclusion-exclusion principle ∣A∪B∣=∣A∣+∣B∣−∣A∩B∣
∣A∪B∣=128+64−32=160\vert{A\cup{}B}\vert=128+64-32=160∣A∪B∣=128+64−32=160
Answer: 160
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