a school cook plans her calendar for the month of february in which there are 20 school days. she plans exactly one meal per school day. unfortunately, she only knows how to cook ten different meals. (a) how many ways are there for her to plan her schedule of menus for the 20 school days if there are no restrictions on the number of times she cooks a particular type of meal?

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16-11-2022
Mathematics

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a school cook plans her calendar for the month of february in which there are 20 school days. she plans exactly one meal per school day. unfortunately, she only knows how to cook ten different meals. (a) how many ways are there for her to plan her schedule of menus for the 20 school days if there are no restrictions on the number of times she cooks a particular type of meal?

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joaobezerra joaobezerra · Answer 1 · 2022-11-23T01:31:00+01:00

The number of ways that there are for her to plan her schedule of menus for the 20 school days, with no restrictions, using the Fundamental Counting Theorem, is of:

10^20.

What is the Fundamental Counting Theorem?

The Fundamental Counting Theorem states that if there are n independent trials, each with [tex]n_1, n_2, \cdots, n_n[/tex] possible results, the total number of outcomes is calculated by the multiplication of the number of outcomes for each trial as presented as follows:

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

In this problem, we have that:

In each day she cooks one meal, hence the number of trials is of 20.

There are no restrictions, and she knows how to cook ten different meals, hence for each trial, the number of possible outcomes is of 10.

Then the number of different meals is obtained as follows:

N = 10^20.

As there are 20 trials, each with 10 possible outcomes.

More can be learned about the Fundamental Counting Theorem at https://brainly.com/question/15878751

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