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Once the first two letters are used, that leaves us 24 letters for the third choice, 23 letters for the fourth choice, 22 letters for the fifth choice, and 21 letters for the sixth choice. (Enter the missing values.)

[tex]\[26 \cdot 25 \cdot 24 \cdot 23 \cdot 22 \cdot 21\][/tex]

Now we need to enter the amount of numbers we have from which to choose. There are 10 numbers: [tex]\(0, 1, 2, 3, 4, 5, 6, 7, 8, 9\)[/tex].

We will enter 10 in the seventh space. Since we cannot reuse the number, we will have a choice of 9 numbers left. (Enter the values below.)

[tex]\[26 \cdot 25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 10 \cdot 9\][/tex]

Step 3: Now multiply the values to get [tex]\(14,918,904,000\)[/tex]. That is almost 15 billion passwords.

How many passwords would the website have if users were allowed to reuse the letters and numbers? (Enter the number of successful outcomes in each of the blanks below. Let the first six spaces represent letters and the last two spaces represent numbers.)

[tex]\[
\text{Letters: } 26 \cdot 26 \cdot 26 \cdot 26 \cdot 26 \cdot 26
\][/tex]

[tex]\[
\text{Numbers: } 10 \cdot 10
\][/tex]

That is almost [tex]\(30,891,577,600\)[/tex] passwords!

That is almost 30.9 billion passwords!


Sagot :

Sure, let's walk through the problem step by step.

### Without Reusing Characters
1. Using Letters Without Reuse:
- 26 options for the first letter.
- 25 options for the second, since one letter has been used.
- 24 options for the third, since two letters have been used.
- 23 options for the fourth, since three letters have been used.
- 22 options for the fifth, since four letters have been used.
- 21 options for the sixth, since five letters have been used.

So, the calculation for the six letters without reuse looks like this:
[tex]$ 26 \times 25 \times 24 \times 23 \times 22 \times 21 $[/tex]

2. Using Numbers Without Reuse:
- 10 options for the first number.
- 9 options for the second number, since one number has been used.

So, the calculation for the two numbers without reuse looks like this:
[tex]$ 10 \times 9 $[/tex]

3. Combining Both:
Combining the calculations for letters and numbers without reuse:
[tex]$ 26 \times 25 \times 24 \times 23 \times 22 \times 21 \times 10 \times 9 = 14,918,904,000 $[/tex]

So, the total number of unique passwords without reuse is 14,918,904,000.

### With Reusing Characters
1. Using Letters With Reuse:
- For each position, you have all 26 letters available, as characters can be reused.

Thus, for six letters:
[tex]$ 26 \times 26 \times 26 \times 26 \times 26 \times 26 = 26^6 $[/tex]

2. Using Numbers With Reuse:
- For each position, you have all 10 numbers available, as numbers can be reused.

Thus, for two numbers:
[tex]$ 10 \times 10 = 10^2 $[/tex]

3. Combining Both:
Combining the calculations for letters and numbers with reuse:
[tex]$ 26^6 \times 10^2 $[/tex]

Which amounts to:
[tex]$ 26^6 \times 10^2 = 30,891,577,600 $[/tex]

So, the total number of possible passwords with reuse is 30,891,577,600.

### Summary
1. Number of unique passwords without reusing any character: 14,918,904,000.
2. Number of possible passwords with reuse of characters: 30,891,577,600.

That’s almost 30.9 billion passwords!
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