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

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==Solution 2==
 
==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)=<math>\boxed{\textbf{(C)}\ 190}</math>~heeeeeeheeeeeee
 
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)=<math>\boxed{\textbf{(C)}\ 190}</math>~heeeeeeheeeeeee
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{{AMC8 box|year=2019|num-b=6|num-a=26}}

Revision as of 12:23, 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 people has at least $2$ 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



2019 AMC 8 (Problems • Answer Key • Resources)
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
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