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Difference between revisions of "2019 AMC 8 Problems/Problem 25"

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==Problem 25==
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==Problem==
 
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?
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<math>\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380</math>
 
<math>\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380</math>
  
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==Solution 1==
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==Solution 1 (Stars and Bars/Sticks and Stones)==
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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.
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<cmath>\implies k + r + y = 24</cmath>We can manipulate this into an equation which can be solved using stars and bars.
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Note: This solution uses the non-negative version for stars and bars. A solution using the positive version of stars is similar (first removing an apple from each person instead of 2).
  
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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>.
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This method uses the counting method of stars and bars (non-negative version). Since each person must have at least <math>2</math> apples, we can remove <math>2*3</math> apples from the total that need to be sorted. With the remaining <math>18</math> apples, we can use stars and bars to determine the number of possibilities. Assume there are <math>18</math> stars in a row, and <math>2</math> bars, which will be placed to separate the stars into groups of <math>3</math>. In total, there are <math>18</math> spaces for stars <math>+ 2</math> spaces for bars, for a total of <math>20</math> spaces. We can now do <math>20 \choose 2</math>. This is because if we choose distinct <math>2</math> spots for the bars to be placed, each combo of <math>3</math> groups will be different, and all apples will add up to <math>18</math>. We can also do this because the apples are indistinguishable. <math>20 \choose 2</math> is <math>190</math>, therefore the answer is <math>\boxed{\textbf{(C) }190}</math>.
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<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>.
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<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.
 
  
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~goofytaipan91
  
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By Stars and Bars, our answer is just <math>\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{\textbf{(C)}\ 190}</math>.
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==Solution 2 (Answer Choices)==
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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>.  
  
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==Solution 2==
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-BorealBear
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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==
 
==Solution 3==
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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>.  
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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.
  
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- aops5234
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~[[User:Bloggish|Bloggish]]
  
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==Solution 4==
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==Video Solution by Math-X (Let's review stars and bars together first!!!)==
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https://youtu.be/IgpayYB48C4?si=SzBgzW4jHelkYwP1&t=8105
  
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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.
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~Math-X
  
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~by sakshamsethi
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== Video Solution by OmegaLearn ==
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https://youtu.be/5UojVH4Cqqs?t=5131
  
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==Solution 5==
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~ pi_is_3.14
  
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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.
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==Video Solution by The Power of Logic(1 to 25 Full Solution)==
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https://youtu.be/Xm4ZGND9WoY
  
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~by Dolphindesigner
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~Hayabusa1
  
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==Solution 6 (Answer Choices)==
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==Video Solutions==
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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
 
  
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==Video Solutions==
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https://www.youtube.com/watch?v=EJzSOPXULBc
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https://www.youtube.com/watch?v=OPFJ-d1byw4 Math is cool
 
  
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https://www.youtube.com/watch?v=EJzSOPXULBc - Happytwin
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- Happytwin
  
 
https://www.youtube.com/watch?v=wJ7uvypbB28
 
https://www.youtube.com/watch?v=wJ7uvypbB28
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https://www.youtube.com/watch?v=2dBUklyUaNI
 
https://www.youtube.com/watch?v=2dBUklyUaNI
  
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https://youtu.be/ZsCRGK4VgBE ~DSA_Catachu
 
  
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https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7 ~ MathEx
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https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7  
  
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https://youtu.be/5UojVH4Cqqs?t=5131 ~ pi_is_3.14
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~ MathEx
  
 
https://youtu.be/8kzjB60pBrA
 
https://youtu.be/8kzjB60pBrA
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~savannahsolver
 
~savannahsolver
  
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==See Also==
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==See also==
 
{{AMC8 box|year=2019|num-b=24|after=Last Problem}}
 
{{AMC8 box|year=2019|num-b=24|after=Last Problem}}
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[[Category:Introductory Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Combinatorics Problems]]
 

Latest revision as of 09:33, 9 November 2024

Problem

Alice has $24$ 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?

$\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$

Solution 1 (Stars and Bars/Sticks and Stones)

Note: This solution uses the non-negative version for stars and bars. A solution using the positive version of stars is similar (first removing an apple from each person instead of 2).

This method uses the counting method of stars and bars (non-negative version). Since each person must have at least $2$ apples, we can remove $2*3$ apples from the total that need to be sorted. With the remaining $18$ apples, we can use stars and bars to determine the number of possibilities. Assume there are $18$ stars in a row, and $2$ bars, which will be placed to separate the stars into groups of $3$. In total, there are $18$ spaces for stars $+ 2$ spaces for bars, for a total of $20$ spaces. We can now do $20 \choose 2$. This is because if we choose distinct $2$ spots for the bars to be placed, each combo of $3$ groups will be different, and all apples will add up to $18$. We can also do this because the apples are indistinguishable. $20 \choose 2$ is $190$, therefore the answer is $\boxed{\textbf{(C) }190}$.


~goofytaipan91

Solution 2 (Answer Choices)

Consider an unordered triple $(a,b,c)$ where $a+b+c=24$ and $a,b,c$ are not necessarily distinct. Then, we will either have $1$, $3$, or $6$ distinguishable ways to assign $a$, $b$, and $c$ to Alice, Becky, and Chris. Thus, our answer will be $x+3y+6z$ for some nonnegative integers $x,y,z$. Notice that we only have $1$ way to assign the numbers $a,b,c$ to Alice, Becky, and Chris when $a=b=c$. As this only happens $1$ way ($a=b=c=8$), our answer is $1+3y+6z$ for some $y,z$. Finally, notice that this implies the answer is $1$ mod $3$. The only answer choice that satisfies this is $\boxed{\textbf{(C) }190}$.

-BorealBear

Solution 3

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

~Bloggish

Video Solution by Math-X (Let's review stars and bars together first!!!)

https://youtu.be/IgpayYB48C4?si=SzBgzW4jHelkYwP1&t=8105

~Math-X

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution by The Power of Logic(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

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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