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"

(→‎Solution 1)
(→‎Solution 1: improve clarity, rigor of solution 1)
Line 3: Line 3:
  
 
==Solution 1==
 
==Solution 1==
−
It is easier to use [[Stars and bars]] when all the numbers are nonnegative, rather than <math>\geq 2</math>. So we redefine variables so that the sum is <math>24-6</math> and each number is nonnegative. Using <math>18</math> apples and <math>2</math> bars (to split it up into <math>3</math> parts), we get <math>{20 \choose 2}</math>, which is equal to <math>\boxed{\textbf{(C) }190}</math>.
+
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>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 17:52, 25 November 2019

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?

Solution 1

We use stars and bars. The problem asks for the number of integer solutions $(a,b,c)$ such that $a+b+c = 24$ and $a,b,c \ge 2$. We can subtract 2 from $a$, $b$, $c$, so that we equivalently seek the number of non-negative integer solutions to $a' + b' + c' = 18$. By stars and bars (using 18 stars and 2 bars), the number of solutions is $\binom{18+2}{2} = \binom{20}{2} = \boxed{\textbf{(C) }190}$.

Solution 2

Let's say you 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 (19/2) = $\boxed{\textbf{(C)}\ 190}$

~heeeeeeheeeeeee

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=112183"