2019 AMC 8 Problems/Problem 25

Revision as of 22:38, 24 November 2019 by Go1331 (talk | contribs) (Solution 1)

Problem 25

Alice has $24$ 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 $\geq 2$. So we redefine variables so that the sum is $24-6$ and each number is nonnegative. Using $18$ apples and $2$ bars (to split it up into $3$ parts), we get ${20 \choose 2}$, which is equal to $\boxed{\textbf{(C) }190}$.

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) = $\boxed{\textbf{(C)}\ 190}$

~heeeeeeheeeeeee

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
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All AJHSME/AMC 8 Problems and Solutions

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