Difference between revisions of "2019 AMC 8 Problems/Problem 25"
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<math>\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380</math> | <math>\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380</math> | ||
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We use [[stars and bars]]. The problem asks for the number of integer solutions <math>(a,b,c)</math> such that <math>a+b+c = 24</math> and <math>a,b,c \ge 2</math>. We can subtract 2 from <math>a</math>, <math>b</math>, <math>c</math>, so that we equivalently seek the number of non-negative integer solutions to <math>a' + b' + c' = 18</math>. By stars and bars (using 18 stars and 2 bars), the number of solutions is <math>\binom{18+2}{2} = \binom{20}{2} = \boxed{\textbf{(C) }190}</math>. | We use [[stars and bars]]. The problem asks for the number of integer solutions <math>(a,b,c)</math> such that <math>a+b+c = 24</math> and <math>a,b,c \ge 2</math>. We can subtract 2 from <math>a</math>, <math>b</math>, <math>c</math>, so that we equivalently seek the number of non-negative integer solutions to <math>a' + b' + c' = 18</math>. By stars and bars (using 18 stars and 2 bars), the number of solutions is <math>\binom{18+2}{2} = \binom{20}{2} = \boxed{\textbf{(C) }190}</math>. | ||
Revision as of 20:24, 26 January 2020
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
We use stars and bars. The problem asks for the number of integer solutions 
such that 
and 
. We can subtract 2 from 
, 
, 
, so that we equivalently seek the number of non-negative integer solutions to 
. By stars and bars (using 18 stars and 2 bars), the number of solutions is 
.
Solution 2
Without loss of generality, let's assume that Alice has 
apples. There are 
ways to split the rest of the apples with Becky and Chris. If Alice has 
apples, there are 
ways to split the rest of the apples with Becky and Chris. If Alice has 
apples, there are 
ways to split the rest. So the total number of ways to split apples between the three friends is equal to 
~heeeeeeheee
Solution 3
Let's assume that the three of them have 
apples. Since each of them has to have at least apples, we say that 
and 
. Thus, 
, and so by stars and bars, the number of solutions for this is 
- aops5234
Video Solution
https://www.youtube.com/watch?v=2dBUklyUaNI
Community discussions
https://artofproblemsolving.com/community/c5h1955387
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
