# 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 (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> distinguishable 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 | ||

+ | |||

+ | ==Solution 3== | ||

+ | Since each person needs to have at least two apples, we can simply give each person two, leaving <math> 24 - 2\times3=18 </math> apples. For the remaining apples, if Alice is going to have <math> a </math> apples, Becky is going to have <math> b </math> apples, and Chris is going to have <math> c </math> apples, we have indeterminate equation <math> a+b+c=18 </math>. Currently, we can see that <math> 0 \leq a\leq 18 </math> where <math> a </math> is an integer, and when <math> a </math> equals any number in the range, there will be <math> 18-a+1=19-a </math> sets of values for <math> b </math> and <math> c </math>. Thus, there are <math> 19 + 18 + 17 + \cdots + 1 = \boxed{\textbf{(C) }190} </math> possible sets of values in total. | ||

+ | |||

+ | ~[[User:Bloggish|Bloggish]] | ||

+ | |||

+ | == Video Solution by OmegaLearn == | ||

+ | https://youtu.be/5UojVH4Cqqs?t=5131 | ||

+ | |||

+ | ~ pi_is_3.14 | ||

+ | |||

+ | ==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)== | ||

+ | https://youtu.be/Xm4ZGND9WoY | ||

+ | |||

+ | ~Hayabusa1 | ||

+ | |||

+ | ==Video Solutions== | ||

+ | |||

+ | https://www.youtube.com/watch?v=EJzSOPXULBc | ||

+ | |||

+ | - Happytwin | ||

+ | |||

+ | https://www.youtube.com/watch?v=wJ7uvypbB28 | ||

+ | |||

+ | https://www.youtube.com/watch?v=2dBUklyUaNI | ||

+ | |||

+ | |||

+ | https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7 | ||

+ | |||

+ | ~ MathEx | ||

+ | |||

+ | https://youtu.be/8kzjB60pBrA | ||

+ | |||

+ | ~savannahsolver | ||

==See Also== | ==See Also== | ||

− | |||

{{MAA Notice}} | {{MAA Notice}} | ||

+ | |||

+ | [[Category:Introductory Combinatorics Problems]] |

## Latest revision as of 18:49, 6 August 2023

## 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 (Answer Choices)

Consider an unordered triple where and are not necessarily distinct. Then, we will either have , , or distinguishable 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

## Solution 3

Since each person needs to have at least two apples, we can simply give each person two, leaving apples. For the remaining apples, if Alice is going to have apples, Becky is going to have apples, and Chris is going to have apples, we have indeterminate equation . Currently, we can see that where is an integer, and when equals any number in the range, there will be sets of values for and . Thus, there are possible sets of values in total.

## Video Solution by OmegaLearn

https://youtu.be/5UojVH4Cqqs?t=5131

~ pi_is_3.14

## Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)

~Hayabusa1

## Video Solutions

https://www.youtube.com/watch?v=EJzSOPXULBc

- Happytwin

https://www.youtube.com/watch?v=wJ7uvypbB28

https://www.youtube.com/watch?v=2dBUklyUaNI

https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7

~ MathEx

~savannahsolver

## See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.