Difference between revisions of "2019 AMC 8 Problems/Problem 25"
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==Solution 1== | ==Solution 1== | ||
| − | 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 |
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| + | ==See Also== | ||
| + | {{AMC8 box|year=2019|num-b=1|num-a=3}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 02:28, 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
See Also
| 2019 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
