Eight people are going for a ride in a boat that seats eight people. One person will drive, and only three of the remaining people are willing to ride in the two bow seats. How many seating arrangements are possible?

the combination of the five seats has 5! times 2 combinations for each of the 3 passengers willing to ride in the two boat seats thus the total number of different seating arrangements is 5! times 3! or 720

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joaobezerra joaobezerra · Answer 2 · 2022-03-18T13:28:34+01:00

Using the Fundamental Counting Theorem, it is found that there are 5760 possible seating arrangements.

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]

In this problem:

For the driver, there are 8 outcomes, hence [tex]n_1 = 8[/tex].

For the bow seats, there are [tex]n_2 = 3 \times 2 = 6[/tex] outcomes.

For the other 5 seats, there are [tex]n_3 = 5![/tex] possible outcomes.

Hence:

[tex]N = 8 \times 6 \times 5! = 5760[/tex]

There are 5760 possible seating arrangements.

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