# Difference between revisions of "2019 AMC 8 Problems/Problem 25"

## 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

 2019 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast 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