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

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

Revision as of 15:20, 28 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

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) = $\boxed{\textbf{(C)}\ 190}$

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

See Also

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

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