Answered

A company prints contest codes on its fun-size bags of
candy. Each 6-character code consists of the letter A
followed by the letter H followed by 4 of the digits 0
through 9. The digits may repeat. Which of the
following expressions gives the number of different
6-character codes that are possible?
F. 1(1)(10)(10)(10) (10)
G. 2(1)(10)(9)(8)(7)
H. 2(1)(10)(10)(10)(10)
J. 2(2)(10)(9)(8)(7)
K. 2(2)(10)(10)(10)(10)

Answer :

Using the Fundamental Counting Theorem, it is found that the expression that  gives the number of different  6 -character codes that are possible is:

F. 1(1)(10)(10)(10) (10)

Fundamental counting theorem:

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]

In this problem:

  • For the first two digits, there is only one outcome, A and H, respectively, hence [tex]n_1 = n_2 = 1[/tex].
  • For the next four digits, as the digits may repeat, there are ten possible outcomes for each, hence [tex]n_3 = n_4 = n_5 = n_6[/tex].

Hence:

[tex]N = 1 \times 1 \times 10 \times 10 \times 10 \times 10[/tex]

And option F is correct.

To learn more about the Fundamental Counting Theorem, you can take a look at https://brainly.com/question/24314866

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