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

(Solution 2)
(Fixed Problem statement)
Line 1: Line 1:
 
==Problem 25==
 
==Problem 25==
Alice has <math>24</math> apples. In how many ways can she share them with Becky and Chris so that each of the people has at least <math>2</math> apples?
+
Alice has <math>24</math> 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==
 
==Solution 1==

Revision as of 12:33, 20 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

Using Stars and bars, and removing $6$ apples so each person can have $2$, we get the total number of ways, which is ${20 \choose 2}$, which is equal to $\boxed{\textbf{(C) }190}$. ~~SmileKat32

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 26
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