Question

A world-famous prix fixe menu includes the following choices.

Beverage (choose one): wine, soft drink, coffee, tea, milk, juice, or cappuccino

First course (no choice): baguette, bruschetta, whole roasted garlic, and butter

Second course (choose one): quesadilla, onion soup, salad, Caesar salad, lettuce wedge, spinach salad, or romaine salad

Third course (choose one): rib eye steak, swordfish, rack of lamb, whitefish, grilled chicken, southern fried chicken, lobster, or vegetarian plate
if you choose steak, then choose one of the following: fries, mashed potatoes, creamed corn, spinach, broccoli, or grilled onions

Fourth course (choose one): chocolate cake, cinnamon tart, apple tart, pecan tart, creme brulee, mango sorbet, or chocolate tart

How many different choices are possible?

Answer 1

Using the fundamental counting theorem, it is found that 107,016 different choices are possible.

Fundamental counting theorem:

States that if there are n things, each with $n_1, n_2, …, n_n$ ways to be done, each thing independent of the other, the number of ways they can be done is:

$N = n_1 \times n_2 \times \cdots \times n_n$

In this problem:

• For the beverage, there are 7 options, hence $n_1 = 7$

• For the first course, there are no choice, hence an arrangement of 4 options, that is, $n_2 = 4! = 24$

• For the second course, there are 7 options, hence $n_3 = 7$

• For the third course, there are 7 non-steak options, plus 6 steak options, hence $n_4 = 13$
• For the fourth course, there are 7 choices, hence $n_4 = 7$

Then, applying the theorem:

$N = 7(24)(7)(13)(7) = 107016$

107,016 different choices are possible.

A similar problem is given at https://brainly.com/question/19022577