parton98
Answered

A password is 4 characters long and must consist of 3 letters and 1 of 10 special characters. If letters can be repeated and the special character is at the end of the password, how many possibilities are there?


A175,760
B456,976
C703,040
D1,679,616

Verified Answer :

caylus

Hello,
number of letters:26
In each 3 first position ,we can put 26 possibilities
=>26^3
Last position: 10 posibilities

==>26^3*10 =175 760
Answer A

Using the Fundamental Counting Theorem, it is found that the number of possibilities is given by:

a. 175,760

What is the Fundamental Counting Theorem?

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

\(N = n_1 \times n_2 \times \cdots \times n_n\)

In this problem:

  • The first three characters are letters that can be repeated, hence \(n_1 = n_2 = n_3 = 26\).
  • The last character is one digit from a set of 10, hence \(n_4 = 10\).

Then:

N = 26 x 26 x 26 x 10 = 175,760.

Hence option A is correct.

To learn more about the Fundamental Counting Theorem, you can check https://brainly.com/question/24314866

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