Answer to Question #155155 in Combinatorics | Number Theory for maryam

Question

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 

Answer 1

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