Question

At a local restaurant, a new dinner menu is available that offers 4 courses — an appetizer, a salad, a main dish, and a dessert — all for one price. There are 5 appetizers, 2 salads, 3 main dishes, and 5 desserts. How many dinner outcomes are possible?

Answer 1

so the answer would be 150 all you have to do it multiply all of them to get 150

Answer 2

The number of possible dinner outcomes = 150

What is combination?

"It is a way of selecting items from a collection "

What is formula of combination?

"$^nC_r=\frac{n!}{r!(n-r)!}$"

What is n factorial?

n! = n × (n - 1) × . . . × 2 × 1

For given question,

A new dinner menu is available that offers 4 courses — an appetizer, a salad, a main dish, and a dessert.

There are 5 appetizers, 2 salads, 3 main dishes, and 5 desserts.

Using combination formula the number of possible dinner outcomes would be,

$^5C_1\times ^2C_1\times ^3C_1\times ^5C_1\\\\=\frac{5!}{1!(5-1)!}\times \frac{2!}{1!(2-1)!}\times \frac{3!}{1!(3-1)!}\times \frac{5!}{1!(5-1)!} \\\\=\frac{5\times 4!}{4!}\times \frac{2 \times 1!}{1!}\times \frac{3\times 2!}{2!}\times \frac{5\times 4!}{4!} \\\\=5\times 2\times 3\times 5\\\\=150$

Therefore, the number of possible dinner outcomes = 150

Learn more about combination here:

brainly.com/question/13387529

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