Difference between revisions of "2001 AMC 10 Problems/Problem 19"
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Let's use [[stars and bars]]. | Let's use [[stars and bars]]. | ||
Let the donuts be represented by <math> O </math>s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us <math> 4 </math> in all. The four donuts we want can be represented as <math> OOOO </math>. Notice that we can add two "dividers" to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, <math> O|OO|O </math> represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in <math> \binom{6}{2}=15 </math> ways. Our answer is hence <math> \boxed{\textbf{(D)}\ 15} </math>. Notice that this can be generalized to get the balls and urn (stars and bars) identity. | Let the donuts be represented by <math> O </math>s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us <math> 4 </math> in all. The four donuts we want can be represented as <math> OOOO </math>. Notice that we can add two "dividers" to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, <math> O|OO|O </math> represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in <math> \binom{6}{2}=15 </math> ways. Our answer is hence <math> \boxed{\textbf{(D)}\ 15} </math>. Notice that this can be generalized to get the balls and urn (stars and bars) identity. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Simple casework works here as well: | ||
+ | Set up the following ratios: | ||
+ | 4:0:0 | ||
+ | 3:1:0 | ||
+ | 2:2:0 | ||
+ | 2:1:1 | ||
+ | |||
+ | In three of these cases we see that there are two of the same ratios (so like two boxes would have 0), and so if we swapped those two donuts, we would have the same case. Thus we get <math>\frac{4!}{3!2!}</math> for those 3 (You can also set it up and logically symmetry applies). For the other case where each ratio of donuts is different, we get the normal 4C3 = 6. Thus, our answer is <math>3*3+6 = \boxed{15}</math> | ||
== See Also == | == See Also == |
Revision as of 21:49, 8 January 2019
Contents
Problem
Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?
Solution
Let's use stars and bars. Let the donuts be represented by s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us in all. The four donuts we want can be represented as . Notice that we can add two "dividers" to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in ways. Our answer is hence . Notice that this can be generalized to get the balls and urn (stars and bars) identity.
Solution 2
Simple casework works here as well: Set up the following ratios: 4:0:0 3:1:0 2:2:0 2:1:1
In three of these cases we see that there are two of the same ratios (so like two boxes would have 0), and so if we swapped those two donuts, we would have the same case. Thus we get for those 3 (You can also set it up and logically symmetry applies). For the other case where each ratio of donuts is different, we get the normal 4C3 = 6. Thus, our answer is
See Also
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 AMC 10 Problems and Solutions |
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