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

(Solution 2)
(Solution 2)
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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 =  
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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>
20(19/2)=<math>\boxed{\textbf{(C)}\ 190}</math>~heeeeeeheeeeeee
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~heeeeeeheeeeeee
  
 
==See Also==
 
==See Also==

Revision as of 14:41, 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

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