2001 AMC 8 Problems/Problem 14

Problem

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

$\text{(A)}\ 4 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 72 \qquad \text{(D)}\ 80 \qquad \text{(E)}\ 144$

Solution

There are $3$ possibilities for the meat and $4$ possibilites for the dessert, for a total of $4\times3=12$ possibilities for the meat and the dessert. There are $4$ possibilities for the first vegetable and $3$ possibilities for the second, but order doesn't matter, so we overcounted by a factor of $2$ . For example, we counted 'baked beans and corn' and 'corn and baked beans' as $2$ different possibilities, so the total possibilites for the two vegetables is $\frac{4\times3}{2}=6$ , and the total number of possibilites is $12\times6=72, \boxed{\text{C}}$

2001 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

Page

Toolbox

Search

2001 AMC 8 Problems/Problem 14

Problem

Solution

See Also