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
[hide]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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