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

Revision as of 06:06, 8 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 $\binom{21-1}{3-1}=\binom{20}{2}=\boxed{\textbf{(C)}\ 190}$ .

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 4: / Line 4: @@
 ==Solution 1==
-We use [[stars and bars]]. The problem asks for the number of integer solutions <math>(a,b,c)</math> such that <math>a+b+c = 24</math> and <math>a,b,c \ge 2</math>. We can subtract 2 from <math>a</math>, <math>b</math>, <math>c</math>, so that we equivalently seek the number of non-negative integer solutions to <math>a' + b' + c' = 18</math>. By stars and bars (using 18 stars and 2 bars), the number of solutions is <math>\binom{18+2}{2} = \binom{20}{2} = \boxed{\textbf{(C) }190}</math>.
+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.
+<cmath>\implies k + r + y = 24</cmath>We can manipulate this into an equation which can be solved using stars and bars.
+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>.
+<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>.
+<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.
+By Stars and Bars, our answer is just <math>\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{190}</math>.
 ==Solution 2==

2019 AMC 8 (Problems • Answer Key • Resources)
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