0606 P13 - Nov 2022 - Q9 - 4 marks
7880
A \(6\)-character password is to be formed from the following characters.
Letters: \(A,\ B,\ C,\ D\)
Numbers: \(1,\ 2,\ 3,\ 4\)
Symbols: \(*,\ \#,\ \$,\ \pounds\)
No character may be used more than once in any password.
(a)(i) Find the number of different \(6\)-character passwords that can be formed.
(a)(ii) How many of these \(6\)-character passwords end with a symbol?
(b) Find the number of different \(6\)-character passwords that include all the symbols, but do not start or end with a symbol.
Solution
Mark Scheme
Solution
Answer: (a)(i) \(665280\); (a)(ii) \(221760\); (b) \(1344\).
Answer: (a)(i) \(665280\); (a)(ii) \(221760\); (b) \(1344\).
There are \(12\) different characters in total.
(a)(i) A \(6\)-character password with no repeated character can be formed in
\({}^{12}P_6\)
ways. Therefore the number is
\(12\times11\times10\times9\times8\times7=665280.\)
(a)(ii) If the password ends with a symbol, there are \(4\) choices for the final character. The first five characters are chosen and arranged from the remaining \(11\) characters:
\(4\times{}^{11}P_5.\)
So the number is
\(4\times11\times10\times9\times8\times7=221760.\)
(b) The password must include all \(4\) symbols and must not start or end with a symbol.
So the first and last positions must be filled by two non-symbol characters. There are
\(8\times7\)
ways to choose and arrange these two end characters.
The four symbols then fill the four middle positions in
\(4!\)
ways. Therefore the number of passwords is
\(8\times7\times4!=8\times7\times24=1344.\)
