2019 AMC 8 Problems/Problem 25

Revision as of 18:52, 25 November 2019 by Scrabbler94 (talk | contribs) (Solution 1: improve clarity, rigor of 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

We use stars and bars. The problem asks for the number of integer solutions $(a,b,c)$ such that $a+b+c = 24$ and $a,b,c \ge 2$. We can subtract 2 from $a$, $b$, $c$, so that we equivalently seek the number of non-negative integer solutions to $a' + b' + c' = 18$. By stars and bars (using 18 stars and 2 bars), the number of solutions is $\binom{18+2}{2} = \binom{20}{2} = \boxed{\textbf{(C) }190}$.

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
Last Problem
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All AJHSME/AMC 8 Problems and Solutions

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