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 people has at least <math>2</math> apples?
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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>
  
 
==Solution 1==
 
==Solution 1==
Using [[Stars and bars]], and removing <math>6</math> apples so each person can have <math>2</math>, we get the total number of ways, which is <math>{20 \choose 2}</math>, which is equal to <math>\boxed{\textbf{(C) }190}</math>. ~~SmileKat32
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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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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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<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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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==
Let's say you assume that Alice has 2 apples. There are 19 ways to split the rest of the apples with Becky and Chris. If Alice has 3 apples, there are 18 ways to split the rest of the apples with Becky and Chris. If Alice has 4 apples, there are 17 ways to split the rest. So the total number of ways to split 24 apples between the three friends is equal to 19+18+17...……+1=20(19/2)=<math>\boxed{\textbf{(C)}\ 190}</math>
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[[Without loss of generality]], let's 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>
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==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} \implies {18+3-1 \choose 3-1} \implies {20 \choose 2}  = \boxed{\textbf{(C)}\ 190}</math> - aops5234
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==Solution 4==
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We can give each person one apple first so that <math>21</math> 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 <math>\binom{21-1}{3-1}=\binom{20}{2}=\boxed{\textbf{(C)}\ 190}</math>.
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==Solution 5==
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Since we are giving each person at least <math>2</math> apples anyway, lets put <math>2*3 = 6</math> aside. Now we have <math>24-6 = 18</math> more apples to distribute to everyone. Since the apples are all indistinguishable, we can put <math>2</math> "dividers" in between the <math>18</math> apples. Ex. <math>oooo|oooooooo|oooooo</math> There are <math>\binom{18+2}{2} = \boxed{\textbf{(C)}\ 190}</math> ways. -SigmaPiE
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==Videos explaining solution==
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https://www.youtube.com/watch?v=2dBUklyUaNI
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https://www.youtube.com/watch?v=EJzSOPXULBc
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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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==See Also==
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{{AMC8 box|year=2019|num-b=24|after=Last Problem}}
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{{MAA Notice}}

Revision as of 17:42, 22 July 2020

Problem 25

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

We use stars and bars. Let Alice get $k$ apples, let Becky get $r$ apples, let Chris get $y$ apples. \[\implies k + r + y = 24\]We can manipulate this into an equation which can be solved using stars and bars.

All of them get at least $2$ apples, so we can subtract $2$ from $k$, $2$ from $r$, and $2$ from $y$. \[\implies (k - 2) + (r - 2) + (y - 2) = 18\]Let $k' = k - 2$, let $r' = r - 2$, let $y' = y - 2$. \[\implies k' + r' + y' = 18\]We can allow either of them to equal to $0$, hence this can be solved by stars and bars.


By Stars and Bars, our answer is just $\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{190}$.

Solution 2

Without loss of generality, let's assume that Alice has $2$ apples. There are $19$ ways to split the rest of the apples with Becky and Chris. If Alice has $3$ apples, there are $18$ ways to split the rest of the apples with Becky and Chris. If Alice has $4$ apples, there are $17$ ways to split the rest. So the total number of ways to split $24$ apples between the three friends is equal to $19 + 18 + 17...…… + 1 = 20\times \frac{19}{2}=\boxed{\textbf{(C)}\ 190}$

Solution 3

Let's assume that the three of them have $x, y, z$ apples. Since each of them has to have at least $2$ apples, we say that $a+2=x, b+2=y$ and $c+2=z$. Thus, $a+b+c+6=24 \implies a+b+c=18$, and so by stars and bars, the number of solutions for this is ${n+k-1 \choose k} \implies {18+3-1 \choose 3-1} \implies {20 \choose 2}  = \boxed{\textbf{(C)}\ 190}$ - aops5234

Solution 4

We can give each person one apple first so that $21$ 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 $\binom{21-1}{3-1}=\binom{20}{2}=\boxed{\textbf{(C)}\ 190}$.

Solution 5

Since we are giving each person at least $2$ apples anyway, lets put $2*3 = 6$ aside. Now we have $24-6 = 18$ more apples to distribute to everyone. Since the apples are all indistinguishable, we can put $2$ "dividers" in between the $18$ apples. Ex. $oooo|oooooooo|oooooo$ There are $\binom{18+2}{2} = \boxed{\textbf{(C)}\ 190}$ ways. -SigmaPiE

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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All AJHSME/AMC 8 Problems and Solutions

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