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

Revision as of 16:42, 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

@@ Line 23: / Line 23: @@
 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>.
+==Solution 5==
+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
 ==Videos explaining solution==

2019 AMC 8 (Problems • Answer Key • Resources)
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