2001 AMC 8 Problem 14

Attempt Problem 14 of the 2001 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

Tyler has entered a buffet line in which he chooses one kind of meat, two different vegetables and one dessert. If the order of food items is not important, how many different meals might he choose?
Meat: beef, chicken, pork
Vegetables: baked beans, corn, potatoes, tomatoes
Dessert: brownies, chocolate cake, chocolate pudding, ice cream
44
2424
7272
8080
144144
Answer: C
Concepts:combinationsmultiplication principle
Small hint:
Choose the two vegetables as an unordered pair
Big hint:
Multiply meat choices, vegetable pairs, and dessert choices
Solution:
He has 33 choices for the meat and 44 choices for dessert.
He must choose 22 of the 44 vegetables. Since order does not matter, there are 4⋅32=6\dfrac{4\cdot3}{2}=6 vegetable choices.
This gives 3⋅6⋅4=723\cdot6\cdot4=72 possible meals.
Thus, C is the correct answer.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.