Difference between revisions of "2019 AMC 8 Problems/Problem 25"
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+ | ==Solution 3== | ||
+ | Lets's assume that the three of them have <math>x, y, z</math> apples. Since each of them has to have at least <math>2</math> apples, we say that <math>a+2=x, b+2=y</math> and <math>c+2=z</math>. Thus, <math>a+b+c+6=24 \implies a+b+c=18</math>, and so by stars and bars, the number of solutions for this is <math>{18+3-1 \choose 3-1} = {20 \choose 2} = \boxed{\textbf{(C)}\ 190}</math> | ||
==See Also== | ==See Also== |
Revision as of 15:20, 28 November 2019
Problem 25
Alice has apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?
Solution 1
We use stars and bars. The problem asks for the number of integer solutions such that and . We can subtract 2 from , , , so that we equivalently seek the number of non-negative integer solutions to . By stars and bars (using 18 stars and 2 bars), the number of solutions is .
Solution 2
Let's say you assume that Alice has 2 apples. There are 19 ways to split the rest of the apples with Becky and Chris. If Alice has 3 apples, there are 18 ways to split the rest of the apples with Becky and Chris. If Alice has 4 apples, there are 17 ways to split the rest. So the total number of ways to split 24 apples between the three friends is equal to 19 + 18 + 17...…… + 1 = 20 (19/2) =
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Solution 3
Lets's assume that the three of them have apples. Since each of them has to have at least apples, we say that and . Thus, , and so by stars and bars, the number of solutions for this is
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
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