Now multiply the values to get 14,918,904,000. That is almost 15 billion passwords.

That is a lot of 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)

Someone please help me!!



Answer :

Using the Fundamental Counting Theorem, it is found that the website would have 30,891,577,600 passwords.

What is the Fundamental Counting Theorem?

It is a theorem that states that if there are n things, each with [tex]n_1, n_2, \cdots, n_n[/tex] ways to be done, each thing independent of the other, the number of ways they can be done is:

[tex]N = n_1 \times n_2 \times \cdots \times n_n[/tex]

For this problem, the first six spaces of the password are letters, and each can assume 26 values, hence the parameters are:

[tex]n_1 = n_2 = n_3 = n_4 = n_5 = n_6 = 26[/tex]

The last two spaces are composed by digits, and each can assume 10 values, hence the parameters are:

[tex]n_7 = n_8 = 10[/tex]

Hence the number of possible passwords is given by:

N = 26^6 x 10² = 30,891,577,600.

The website would have 30,891,577,600 passwords.

More can be learned about the Fundamental Counting Theorem at https://brainly.com/question/24314866

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