Difference between revisions of "2019 AMC 8 Problems/Problem 25"
Dsa catachu (talk | contribs) (→Videos explaining solution) |
(→Solution 1) |
||
Line 4: | Line 4: | ||
==Solution 1== | ==Solution 1== | ||
− | We use [[stars and bars]]. | + | We use [[stars and bars]]. Let Alice get <math>k</math> apples, let Becky get <math>r</math> apples, let Chris get <math>y</math> apples. |
+ | <cmath>\implies k + r + y = 24</cmath>We can manipulate this into an equation which can be solved using stars and bars. | ||
+ | |||
+ | All of them get at least <math>2</math> apples, so we can subtract <math>2</math> from <math>k</math>, <math>2</math> from <math>r</math>, and <math>2</math> from <math>y</math>. | ||
+ | <cmath>\implies (k - 2) + (r - 2) + (y - 2) = 18</cmath>Let <math>k' = k - 2</math>, let <math>r' = r - 2</math>, let <math>y' = y - 2</math>. | ||
+ | <cmath>\implies k' + r' + y' = 18</cmath>We can allow either of them to equal to <math>0</math>, hence this can be solved by stars and bars. | ||
+ | |||
+ | |||
+ | By Stars and Bars, our answer is just <math>\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{190}</math>. | ||
==Solution 2== | ==Solution 2== |
Revision as of 07:06, 8 July 2020
Contents
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. Let Alice get apples, let Becky get apples, let Chris get apples. We can manipulate this into an equation which can be solved using stars and bars.
All of them get at least apples, so we can subtract from , from , and from . Let , let , let . We can allow either of them to equal to , hence this can be solved by stars and bars.
By Stars and Bars, our answer is just .
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
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
Solution 4
We can give each person one apple first so that apples are shared between the three people, where each person receives at least one apple. Using Stars and Bars, the number of ways to do this is .
Videos explaining solution
https://www.youtube.com/watch?v=2dBUklyUaNI
https://www.youtube.com/watch?v=EJzSOPXULBc
https://youtu.be/ZsCRGK4VgBE ~DSA_Catachu
https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7 ~ MathEx
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.