Difference between revisions of "2019 AMC 8 Problems/Problem 25"
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==Problem 25== | ==Problem 25== | ||
Alice has <math>24</math> 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? | Alice has <math>24</math> 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? | ||
+ | <math>\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380</math> | ||
==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{\textbf{(C)}\ 190}</math>. | ||
==Solution 2== | ==Solution 2== | ||
− | + | First assume that Alice has <math>2</math> apples. There are <math>19</math> ways to split the rest of the apples with Becky and Chris. If Alice has <math>3</math> apples, there are <math>18</math> ways to split the rest of the apples with Becky and Chris. If Alice has <math>4</math> apples, there are <math>17</math> ways to split the rest. So the total number of ways to split <math>24</math> apples between the three friends is equal to <math>19 + 18 + 17...…… + 1 = 20\times \frac{19}{2}=\boxed{\textbf{(C)}\ 190}</math> | |
+ | |||
+ | ==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>{n+k-1 \choose k-1} \implies {18+3-1 \choose 3-1} \implies {20 \choose 2} = \boxed{\textbf{(C)}\ 190}</math> - aops5234 | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Since we have to give each of the <math>3</math> friends at least <math>2</math> apples, we need to spend a total of <math>2+2+2=6</math> apples to solve the restriction. Now we have <math>24-6=18</math> apples left to be divided among Alice, Becky, and Chris, without any constraints. We use the [[Ball-and-urn]] technique, or sometimes known as ([Sticks and Stones]/[Stars and Bars]), to divide the apples. We now have <math>18</math> stones and <math>2</math> sticks, which have a total of <math>\binom{18+2}{2}=\binom{20}{2}=\frac{20\times19}{2} = \boxed{190}</math> ways to arrange. | ||
+ | |||
+ | ~by sakshamsethi | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | Equivalently, we split <math>21</math> apples among <math>3</math> friends with each having at least <math>1</math> apples. We put sticks between apples to split apples into three stacks. So there are 20 spaces to put <math>2</math> sticks. We have <math> \binom{20}{2} = 190</math> different ways to arrange the two sticks. So, there are <math>\boxed{190}</math> ways to split the apples among them. | ||
+ | |||
+ | ~by Dolphindesigner | ||
+ | |||
+ | ==Solution 6 (Answer Choices)== | ||
+ | Consider an unordered triple <math> (a,b,c) </math> where <math> a+b+c=24 </math> and <math> a,b,c </math> are not necessarily distinct. Then, we will either have <math> 1 </math>, <math> 3 </math>, or <math> 6 </math> ways to assign <math> a </math>, <math> b </math>, and <math> c </math> to Alice, Becky, and Chris. Thus, our answer will be <math> x+3y+6z </math> for some nonnegative integers <math> x,y,z </math>. Notice that we only have <math> 1 </math> way to assign the numbers <math> a,b,c </math> to Alice, Becky, and Chris when <math> a=b=c </math>. As this only happens <math> 1 </math> way (<math>a=b=c=8</math>), our answer is <math> 1+3y+6z </math> for some <math> y,z </math>. Finally, notice that this implies the answer is <math> 1 </math> mod <math> 3 </math>. The only answer choice that satisfies this is <math> \boxed{\textbf{(C) }190} </math>. -BorealBear | ||
+ | |||
+ | ==Video Solutions== | ||
+ | https://www.youtube.com/watch?v=OPFJ-d1byw4 Math is cool | ||
+ | |||
+ | https://www.youtube.com/watch?v=EJzSOPXULBc - Happytwin | ||
+ | |||
+ | https://www.youtube.com/watch?v=wJ7uvypbB28 | ||
+ | |||
+ | https://www.youtube.com/watch?v=2dBUklyUaNI | ||
+ | |||
+ | https://youtu.be/ZsCRGK4VgBE ~DSA_Catachu | ||
− | ~ | + | https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7 ~ MathEx |
+ | |||
+ | https://youtu.be/5UojVH4Cqqs?t=5131 ~ pi_is_3.14 | ||
+ | |||
+ | https://youtu.be/8kzjB60pBrA | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] |
Latest revision as of 10:25, 26 April 2022
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
First 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
Since we have to give each of the friends at least apples, we need to spend a total of apples to solve the restriction. Now we have apples left to be divided among Alice, Becky, and Chris, without any constraints. We use the Ball-and-urn technique, or sometimes known as ([Sticks and Stones]/[Stars and Bars]), to divide the apples. We now have stones and sticks, which have a total of ways to arrange.
~by sakshamsethi
Solution 5
Equivalently, we split apples among friends with each having at least apples. We put sticks between apples to split apples into three stacks. So there are 20 spaces to put sticks. We have different ways to arrange the two sticks. So, there are ways to split the apples among them.
~by Dolphindesigner
Solution 6 (Answer Choices)
Consider an unordered triple where and are not necessarily distinct. Then, we will either have , , or ways to assign , , and to Alice, Becky, and Chris. Thus, our answer will be for some nonnegative integers . Notice that we only have way to assign the numbers to Alice, Becky, and Chris when . As this only happens way (), our answer is for some . Finally, notice that this implies the answer is mod . The only answer choice that satisfies this is . -BorealBear
Video Solutions
https://www.youtube.com/watch?v=OPFJ-d1byw4 Math is cool
https://www.youtube.com/watch?v=EJzSOPXULBc - Happytwin
https://www.youtube.com/watch?v=wJ7uvypbB28
https://www.youtube.com/watch?v=2dBUklyUaNI
https://youtu.be/ZsCRGK4VgBE ~DSA_Catachu
https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7 ~ MathEx
https://youtu.be/5UojVH4Cqqs?t=5131 ~ pi_is_3.14
~savannahsolver
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.