Question #155155

you choose one of the five-letter passwords producted by using the A,B,C,D and E,each letter should be used exactly once .what is the maximum number of password that do not have a letter in common with the password you chose (the same letter is not in the same place?)

Hint: if the question was asked for three-letter passwords created using the letters,A,B,and C the answer would be 2 :let the password you choose be ABC .passwords without a common letter:ABC 


1
Expert's answer
2021-01-19T04:30:50-0500

Solution: Derangement of n distinct letters=n!(111!+12!13!+14!15!+.....+(1)nn!)Here, We have given 5 letters as A,B,C,D,EThen derangement of 5 distinct letters as=5!(111!+12!13!+14!15!)=5!(12!13!+14!15!)=(5!2!5!3!+5!4!5!5!)=(5.4.3.2.12.15.4.3.2.13.2.1+5.4.3.2.14.3.2.15.4.3.2.15.4.3.2.1)=[(5.4.3)(5.4)+(5)1]=6020+51=6521=44Hence, 44 is the maximum number of password that do not have a letter in common with the password you chose.Solution: ~Derangement~ of ~n~ distinct~ letters\\=n!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+.....+\frac{(-1)^n}{n!}) \\Here, ~We~ have~ given ~5~ letters~ as ~A,B,C,D,E \\Then ~derangement ~of~ 5~ distinct ~letters ~as \\=5!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}) \\=5!(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}) \\=(\frac{5!}{2!}-\frac{5!}{3!}+\frac{5!}{4!}-\frac{5!}{5!}) \\=(\frac{5.4.3.2.1}{2.1}-\frac{5.4.3.2.1}{3.2.1}+\frac{5.4.3.2.1}{4.3.2.1}-\frac{5.4.3.2.1}{5.4.3.2.1}) \\=[(5.4.3)-(5.4)+(5)-1] \\=60-20+5-1 \\=65-21 \\=44 \\Hence, ~44~ is~ the ~maximum ~number~ of~ password ~that~ do~ not~ have~ a ~letter ~\\in~ common ~with ~the ~password ~you ~chose.


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