Difference between revisions of "2019 AMC 8 Problems/Problem 25"
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==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 variable 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>. ~~SmileKat32 | |
==Solution 2== | ==Solution 2== | ||
Revision as of 21:37, 24 November 2019
Contents
Problem 25
Alice has 
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
It is easier to use Stars and bars when all the numbers are nonnegative, rather than 
. So we redefine variable so that the sum is 
and each number is nonnegative. Using 
apples and 
bars (to split it up into 
parts), we get 
, 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) = 
~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 | ||
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