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

(Solution 1)
(Solution 1)
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==Solution 1==
 
==Solution 1==
It is easier to use [[Stars and bars]] when all the numbers are nonnegative, rather than <math>\geq 2</math>. So we redefine variable so that the sum is <math>24-6</math> and each number is nonnegative. Using <math>18</math> apples and <math>2</math> bars (to split it up into <math>3</math> parts), we get <math>{20 \choose 2}</math>, which is equal to <math>\boxed{\textbf{(C) }190}</math>. ~~SmileKat32
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It is easier to use [[Stars and bars]] when all the numbers are nonnegative, rather than <math>\geq 2</math>. So we redefine variables so that the sum is <math>24-6</math> and each number is nonnegative. Using <math>18</math> apples and <math>2</math> bars (to split it up into <math>3</math> parts), we get <math>{20 \choose 2}</math>, which is equal to <math>\boxed{\textbf{(C) }190}</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 22:38, 24 November 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

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

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
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

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