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? $\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$

Solution 1

We use stars and bars. Let Alice get $k$ apples, let Becky get $r$ apples, let Chris get $y$ apples. $\implies k + r + y = 24$ We can manipulate this into an equation which can be solved using stars and bars.

All of them get at least $2$ apples, so we can subtract $2$ from $k$ , $2$ from $r$ , and $2$ from $y$ . $\implies (k - 2) + (r - 2) + (y - 2) = 18$ Let $k' = k - 2$ , let $r' = r - 2$ , let $y' = y - 2$ . $\implies k' + r' + y' = 18$ We can allow either of them to equal to $0$ ; hence, this can be solved by stars and bars.

By Stars and Bars, our answer is just $\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{\textbf{(C)}\ 190}$ .

Solution 2 (Answer Choices)

Consider an unordered triple $(a,b,c)$ where $a+b+c=24$ and $a,b,c$ are not necessarily distinct. Then, we will either have $1$ , $3$ , or $6$ distinguishable ways to assign $a$ , $b$ , and $c$ to Alice, Becky, and Chris. Thus, our answer will be $x+3y+6z$ for some nonnegative integers $x,y,z$ . Notice that we only have $1$ way to assign the numbers $a,b,c$ to Alice, Becky, and Chris when $a=b=c$ . As this only happens $1$ way ( $a=b=c=8$ ), our answer is $1+3y+6z$ for some $y,z$ . Finally, notice that this implies the answer is $1$ mod $3$ . The only answer choice that satisfies this is $\boxed{\textbf{(C) }190}$ .

-BorealBear

Solution 3

Since each person needs to have at least two apples, we can simply give each person two, leaving $24 - 2\times3=18$ apples. For the remaining apples, if Alice is going to have $a$ apples, Becky is going to have $b$ apples, and Chris is going to have $c$ apples, we have indeterminate equation $a+b+c=18$ . Currently, we can see that $0 \leq a\leq 18$ where $a$ is an integer, and when $a$ equals any number in the range, there will be $18-a+1=19-a$ sets of values for $b$ and $c$ . Thus, there are $19 + 18 + 17 + \cdots + 1 = \boxed{\textbf{(C) }190}$ possible sets of values in total.