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Re: A restaurant menu has several options for tacos. There are 3 [#permalink]
The answer is 180.

3*4*3*5 = 180
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Re: A restaurant menu has several options for tacos. There are 3 [#permalink]
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sandy wrote:
A restaurant menu has several options for tacos. There are 3 types of shells, 4 types of meat, 3 types of cheese, and 5 types of salsa. How many distinct tacos can be ordered assuming that any order contains exactly one of each of the above choices?

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180



Take the task of building a taco and break it into stages.

Stage 1: Select a type of shell
Since there are three types of shells to choose from, we can complete stage 1 in 3 ways

Stage 2: Select a type of meat
Since there are four types of meat to choose from, we can complete stage 2 in 4 ways

Stage 3: Select a type of cheese
Since there are three types of cheese to choose from, we can complete this stage in 3 ways

Stage 4: Select type of salsa
Since there are 5 types of salsa to choose from, we can complete this stage in 5 ways

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus build a taco) in (3)(4)(3)(5) ways (= 180 ways)

Answer: 180

Note: the FCP can be used to solve the MAJORITY of counting questions on the GRE. So, be sure to learn it.

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A restaurant menu has several options for tacos. There are 3 [#permalink]
A restaurant menu has several options for tacos. There are 3 types of shells, 4 types of meat, 3 types of cheese, and 5 types of salsa. How many distinct tacos can be ordered assuming that any order contains exactly one of each of the above choices?

We apply the fundamental counting principle which says that if there are \(n\) ways of doing something, and \(m\) ways of doing another thing after that, then there are \(n ∗ m\) ways to perform both of these actions.

In the above case \(3 * 4 * 3 * 5 = 180\) different types of tacos can be ordered.
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Re: A restaurant menu has several options for tacos. There are 3 [#permalink]
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Re: A restaurant menu has several options for tacos. There are 3 [#permalink]
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