Difference between revisions of "2019 AMC 8 Problems/Problem 25"
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==Solution 1== | ==Solution 1== | ||
− | + | 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>. | |
==Solution 2== | ==Solution 2== |
Revision as of 17:52, 25 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
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
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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