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

Revision as of 14:43, 24 December 2019

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

Without loss of generality, let's 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) =

~heeeeeeheeeeeee ~~small fix by mathboy282

Solution 3

Let'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 ${18+3-1 \choose 3-1} = {20 \choose 2} = \boxed{\textbf{(C)}\ 190}$ - aops5234

Solution 4 (NOT RECOMMENDED!)

List all of the solutions down.

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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