# 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

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

## Solution 3

Lets's assume that the three of them have $x, y, z$ apples. Since each of them has to have at least $2$ apples, we say that $a+2=x, b+2=y$ and $c+2=z$. Thus, $a+b+c+6=24 \implies a+b+c=18$, and so by stars and bars, the number of solutions for this is ${18+3-1 \choose 3-1} = {20 \choose 2} = \boxed{\textbf{(C)}\ 190}$