Difference between revisions of "2019 AMC 8 Problems/Problem 25"
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Using [[Stars and bars]], and removing <math>6</math> apples so each person can have <math>2</math>, we get the total number of ways, which is <math>{20 \choose 2}</math>, which is equal to <math>\boxed{\textbf{(C) }190}</math>. ~~SmileKat32 | Using [[Stars and bars]], and removing <math>6</math> apples so each person can have <math>2</math>, we get the total number of ways, which is <math>{20 \choose 2}</math>, which is equal to <math>\boxed{\textbf{(C) }190}</math>. ~~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)=<math>\boxed{\textbf{(C)}\ 190}</math> | |
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Revision as of 13:19, 20 November 2019
Problem 25
Alice has 
apples. In how many ways can she share them with Becky and Chris so that each of the people has at least 
apples?
Solution 1
Using Stars and bars, and removing 
apples so each person can have , we get the total number of ways, which is 
, which is equal to 
. ~~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)=
