Difference between revisions of "2001 AMC 8 Problems/Problem 14"
(→Solution) |
|||
| (2 intermediate revisions by one other user not shown) | |||
| Line 11: | Line 11: | ||
==Solution== | ==Solution== | ||
| − | There are <math> 3 </math> possibilities for the meat and <math> 4 </math> possibilites for the dessert, for a total of <math> 4\times3=12 </math> possibilities for the meat and the dessert. There are <math> 4 </math> possibilities for the first vegetable and <math> 3 </math> possibilities for the second, but order doesn't matter, so we overcounted by a factor of <math> 2 </math>. For example, we counted 'baked beans and corn' and 'corn and baked beans' as <math> 2 </math> different possibilities, so the total possibilites for the two vegetables is <math> \frac{4\times3}{2}=6 </math>, and the total number of possibilites is <math> 12\times6= | + | There are <math> 3 </math> possibilities for the meat and <math> 4 </math> possibilites for the dessert, for a total of <math> 4\times3=12 </math> possibilities for the meat and the dessert. There are <math> 4 </math> possibilities for the first vegetable and <math> 3 </math> possibilities for the second, but order doesn't matter, so we overcounted by a factor of <math> 2 </math>. For example, we counted 'baked beans and corn' and 'corn and baked beans' as <math> 2 </math> different possibilities, so the total possibilites for the two vegetables is <math> \frac{4\times3}{2}=6 </math>, and the total number of possibilites is <math> 12\times6=\boxed{\textbf{(C)}\ 72}.</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2001|num-b=13|num-a=15}} | {{AMC8 box|year=2001|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:30, 5 October 2022
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
Solution
There are 
possibilities for the meat and 
possibilites for the dessert, for a total of 
possibilities for the meat and the dessert. There are possibilities for the first vegetable and possibilities for the second, but order doesn't matter, so we overcounted by a factor of 
. For example, we counted 'baked beans and corn' and 'corn and baked beans' as different possibilities, so the total possibilites for the two vegetables is 
, and the total number of possibilites is 
See Also
| 2001 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
