Question

1. passwords on a certain web site can have from four to eight characters. a character can be any digit or letter. any password can have at most one digit on this web site. repetitions are allowed.

a) how many four-character passwords are possible?
b) how many eight-character passwords are possible?
c) if a hacker can check one combination every 10 s, how much longer does it take to check all of the eight-character passwords than to check all of the four-character passwords?

1. simplify each expression.

a) \\(\frac{n! + (n - 1)!}{}\\)

Explanation:

Part 7a: Simplify \(\frac{n! + (n - 1)!}{(n - 1)!}\) (assuming the denominator is \((n - 1)!\) as it's a common simplification problem with factorials; if not, the process is similar with the given denominator)

Step 1: Recall the definition of factorial

\(n! = n\times(n - 1)!\)

Step 2: Substitute \(n!\) in the numerator

\(\frac{n\times(n - 1)! + (n - 1)!}{(n - 1)!}\)

Step 3: Factor out \((n - 1)!\) from the numerator

\(\frac{(n - 1)!(n + 1)}{(n - 1)!}\)

Step 4: Cancel out \((n - 1)!\) from numerator and denominator (for \(n

eq1\) to avoid division by zero, but in factorial context \(n\geq2\) usually)
\(n + 1\)

Step 1: Analyze the cases for four - character passwords

We have two cases: passwords with no digits and passwords with exactly one digit.

• Case 1: No digits (all letters)

There are 26 letters (assuming English alphabet, both upper - and lower - case? Wait, the problem says "any letter" – if we consider 26 lowercase or 26 uppercase, but usually in such problems, if not specified, we can consider 26 letters (let's assume 26 letters, and 10 digits). So number of letters \(= 26\), number of digits \(= 10\).
For a password with no digits, each of the 4 characters has 26 choices. So the number of such passwords is \(26^4\) (by the multiplication principle, since for each of the 4 positions, we have 26 options).

• Case 2: Exactly one digit

First, we need to choose the position of the digit. There are \(C(4,1)=\frac{4!}{1!(4 - 1)!}=4\) ways to choose the position of the digit.
The digit has 10 choices, and the remaining \(4 - 1 = 3\) characters are letters, each with 26 choices. So the number of passwords with exactly one digit is \(4\times10\times26^3\) (choose position: 4 ways, digit: 10, letters: \(26^3\)).

Step 2: Calculate the total number of four - character passwords

Total number of four - character passwords \(= 26^4+4\times10\times26^3\)
Factor out \(26^3\):
\(26^3(26 + 4\times10)=26^3(26 + 40)=26^3\times66\)
Calculate \(26^3=26\times26\times26 = 17576\), then \(17576\times66 = 1159016\)

Step 1: Analyze the cases for eight - character passwords

We have two cases: passwords with no digits and passwords with exactly one digit.

• Case 1: No digits (all letters)

Each of the 8 characters has 26 choices. So the number of such passwords is \(26^8\).

• Case 2: Exactly one digit

First, choose the position of the digit. There are \(C(8,1)=\frac{8!}{1!(8 - 1)!}=8\) ways to choose the position of the digit.
The digit has 10 choices, and the remaining \(8 - 1 = 7\) characters are letters, each with 26 choices. So the number of passwords with exactly one digit is \(8\times10\times26^7\).

Step 2: Calculate the total number of eight - character passwords

Total number of eight - character passwords \(= 26^8+8\times10\times26^7\)
Factor out \(26^7\):
\(26^7(26 + 8\times10)=26^7(26 + 80)=26^7\times106\)
Calculate \(26^7 = 26\times26\times26\times26\times26\times26\times26=8031810176\)
Then \(8031810176\times106 = 851371878656\)

Answer:

\(n + 1\)

Part 6a: Four - character passwords