Difference between revisions of "2019 AMC 8 Problems/Problem 25"
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==Solution 3== | ==Solution 3== | ||
− | Let'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} | + | Let'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>{n+k-1 \choose k} \implies {18+3-1 \choose 3-1} \implies {20 \choose 2} = \boxed{\textbf{(C)}\ 190}</math> - aops5234 and DragonWarrior123 |
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==Video Solution== | ==Video Solution== |
Revision as of 18:16, 3 February 2020
Contents
[hide]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
Without loss of generality, let's assume that Alice has apples. There are ways to split the rest of the apples with Becky and Chris. If Alice has apples, there are ways to split the rest of the apples with Becky and Chris. If Alice has apples, there are ways to split the rest. So the total number of ways to split apples between the three friends is equal to
~heeeeeeheee
Solution 3
Let'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 - aops5234 and DragonWarrior123
Video Solution
https://www.youtube.com/watch?v=2dBUklyUaNI
Community discussions
https://artofproblemsolving.com/community/c5h1955387
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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