Chapter 0.4, Problem 19E

(a)

To determine

To find:

The passwords are possible if characters are letters that can be repeated and no character can be repeated.

Answer to Problem 19E

  $1402410240$

Explanation of Solution

Given:

Total character

  $36$

Concept used:

Number of possible password without character repetition is solved with permutation formula

  ${}^n{P}_k=\frac{n!}{\left(n-k\right)!}$

Number of possible password with character repetition is solved with permutation formula

  $n^k$

Calculation:

Using permutation formula for maximum combinations

The solution is

  $\begin{array}{l}=\frac{36!}{\left(36-6\right)!}\\ =\frac{30!\times 31\times 32\times 33\times 34\times 35}{30!}\\ =31\times 32\times 33\times 34\times 35\\ =1402410240\end{array}$

(b)

To determine

To explain:

The type of password is more secure.

Answer to Problem 19E

  $2176782336$

Explanation of Solution

Given:

Total character

  $36$

Concept used:

Number of possible password without character repetition is solved with permutation formula

  ${}^n{P}_k=\frac{n!}{\left(n-k\right)!}$

Number of possible password with character repetition is solved with permutation formula

  $n^k$

Calculation:

Number of possible password with character repetition is solved with permutation formula

  $n^k$

Using the above formula

  $\begin{array}{l}={\left(36\right)}^6\\ =2176782336\end{array}$