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

(Solution 5)
m (Videos explaining solution)
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~by Dolphindesigner
 
~by Dolphindesigner
  
==Videos explaining solution==
+
How i did it 24x23x22=190. BOOM. MIC DROPPED.
  
https://www.youtube.com/watch?v=wJ7uvypbB28
+
https://discord.gg/9Vw73UeF
 
 
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
 
 
 
https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25
 
  
 
==See Also==
 
==See Also==

Revision as of 19:55, 8 November 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 $\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{\textbf{(C)}\ 190}$.

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

Since we have to give each of the $3$ friends at least $2$ apples, we need to spend a total of $2+2+2=6$ apples to solve the restriction. Now we have $24-6=18$ 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 $18$ stones and $2$ sticks, which have a total of $\binom{18+2}{2}=\binom{20}{2}=\frac{20\times19}{2} = \boxed{190}$ ways to arrange.

~by sakshamsethi

Solution 5

Equivalently, we split $21$ apples among $3$ friends with each having at least $1$ apples. We put sticks between apples to split apples into three stacks. So there are 20 spaces to put $2$ sticks. We have $\binom{20}{2} = 190$ different ways to arrange the two sticks. So, there are $\boxed{190}$ ways to split the apples among them.

~by Dolphindesigner

How i did it 24x23x22=190. BOOM. MIC DROPPED.

https://discord.gg/9Vw73UeF

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
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Problem 24
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All AJHSME/AMC 8 Problems and Solutions

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