Solution: Derangement of n distinct letters=n!(1−1!1+2!1−3!1+4!1−5!1+.....+n!(−1)n)Here, We have given 5 letters as A,B,C,D,EThen derangement of 5 distinct letters as=5!(1−1!1+2!1−3!1+4!1−5!1)=5!(2!1−3!1+4!1−5!1)=(2!5!−3!5!+4!5!−5!5!)=(2.15.4.3.2.1−3.2.15.4.3.2.1+4.3.2.15.4.3.2.1−5.4.3.2.15.4.3.2.1)=[(5.4.3)−(5.4)+(5)−1]=60−20+5−1=65−21=44Hence, 44 is the maximum number of password that do not have a letter in common with the password you chose.
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