Question

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?


a. 175,760
b. 456,976
c. 703,040
d. 1,679,616

Answer 1

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

Answer 2

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