2019 AMC 8 Problems/Problem 25

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

First 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 ${n+k-1 \choose k-1} \implies {18+3-1 \choose 3-1} \implies {20 \choose 2}  = \boxed{\textbf{(C)}\ 190}$ - 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

Solution 6 (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 $6$ 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

Video Solutions

https://www.youtube.com/watch?v=OPFJ-d1byw4 Math is cool

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



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



See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS