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
Solution 1
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
Solution 2 (combinatorics)
To find the total ways, we multiply all the combinations of all 3 items:
For meat, it's asking how many ways can you pick item out of or (plain logic.)
For vegetables, it's the amount of ways you can pick items out of or
For dessert, it's the amount of ways you can pick item out of or (also logic).
Therefore, it's
~RandomMathGuy500
See Also
2001 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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