Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎See also)
(→‎See also)
(9 intermediate revisions by 6 users not shown)
Line 3: Line 3:
 
<math>\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380</math>
 
<math>\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380</math>
  
−
==Solution 1==
 
−
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>.
+
==Solution 1 (Stars and Bars/Sticks and Stones)==
−
<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.
 
  
 +
Note: This solution uses the non-negative version for stars and bars. A solution using the positive version of stars is similar (first removing an apple from each person instead of 2).
  
−
By Stars and Bars, our answer is just <math>\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{\textbf{(C)}\ 190}</math>.
+
This method uses the counting method of stars and bars (non-negative version). Since each person must have at least <math>2</math> apples, we can remove <math>2*3</math> apples from the total that need to be sorted. With the remaining <math>18</math> apples, we can use stars and bars to determine the number of possibilities. Assume there are <math>18</math> stars in a row, and <math>2</math> bars, which will be placed to separate the stars into groups of <math>3</math>. In total, there are <math>18</math> spaces for stars <math>+ 2</math> spaces for bars, for a total of <math>20</math> spaces. We can now do <math>20 \choose 2</math>. This is because if we choose distinct <math>2</math> spots for the bars to be placed, each combo of <math>3</math> groups will be different, and all apples will add up to <math>18</math>. We can also do this because the apples are indistinguishable. <math>20 \choose 2</math> is <math>190</math>, therefore the answer is <math>\boxed{\textbf{(C) }190}</math>.
 +
 +
 
 +
~goofytaipan91
  
 
==Solution 2 (Answer Choices)==
 
==Solution 2 (Answer Choices)==
Line 23: Line 22:
  
 
~[[User:Bloggish|Bloggish]]
 
~[[User:Bloggish|Bloggish]]
 +
 +
==Video Solution by Math-X (Let's review stars and bars together first!!!)==
 +
https://youtu.be/IgpayYB48C4?si=SzBgzW4jHelkYwP1&t=8105
 +
 +
~Math-X
  
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==
Line 54: Line 58:
  
 
==See also==
 
==See also==
−
{{AMC8 box|year=2019|num-b=24|num - a = Last Problem}}
+
{{AMC8 box|year=2019|num-b=24|after=Last Problem}}
 +
[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

Revision as of 13:07, 26 June 2024

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 (Stars and Bars/Sticks and Stones)

Note: This solution uses the non-negative version for stars and bars. A solution using the positive version of stars is similar (first removing an apple from each person instead of 2).

This method uses the counting method of stars and bars (non-negative version). Since each person must have at least $2$ apples, we can remove $2*3$ apples from the total that need to be sorted. With the remaining $18$ apples, we can use stars and bars to determine the number of possibilities. Assume there are $18$ stars in a row, and $2$ bars, which will be placed to separate the stars into groups of $3$. In total, there are $18$ spaces for stars $+ 2$ spaces for bars, for a total of $20$ spaces. We can now do $20 \choose 2$. This is because if we choose distinct $2$ spots for the bars to be placed, each combo of $3$ groups will be different, and all apples will add up to $18$. We can also do this because the apples are indistinguishable. $20 \choose 2$ is $190$, therefore the answer is $\boxed{\textbf{(C) }190}$.


~goofytaipan91

Solution 2 (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 $6$ distinguishable 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

Solution 3

Since each person needs to have at least two apples, we can simply give each person two, leaving $24 - 2\times3=18$ apples. For the remaining apples, if Alice is going to have $a$ apples, Becky is going to have $b$ apples, and Chris is going to have $c$ apples, we have indeterminate equation $a+b+c=18$. Currently, we can see that $0 \leq a\leq 18$ where $a$ is an integer, and when $a$ equals any number in the range, there will be $18-a+1=19-a$ sets of values for $b$ and $c$. Thus, there are $19 + 18 + 17 + \cdots + 1 = \boxed{\textbf{(C) }190}$ possible sets of values in total.

~Bloggish

Video Solution by Math-X (Let's review stars and bars together first!!!)

https://youtu.be/IgpayYB48C4?si=SzBgzW4jHelkYwP1&t=8105

~Math-X

Video Solution by OmegaLearn

https://youtu.be/5UojVH4Cqqs?t=5131

~ pi_is_3.14

Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

Video Solutions

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

- Happytwin

https://www.youtube.com/watch?v=wJ7uvypbB28

https://www.youtube.com/watch?v=2dBUklyUaNI


https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7

~ MathEx

https://youtu.be/8kzjB60pBrA

~savannahsolver

See also

2019 AMC 8 (Problems • Answer Key • Resources)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_25&oldid=221128"