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

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==Solution 1==
 
==Solution 1==
Using [[Stars and bars]], and removing <math>6</math> apples so each person can have <math>2</math>, we get the total number of ways, which is <math>{20 \choose 2}</math>, which is equal to <math>\boxed{\textbf{(C) }190}</math>. ~~SmileKat32 (why isn't the answers online?)
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Using [[Stars and bars]], and removing <math>6</math> apples so each person can have <math>2</math>, we get the total number of ways, which is <math>{20 \choose 2}</math>, which is equal to <math>\boxed{\textbf{(C) }190}</math>. ~~SmileKat32  
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==See Also==
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{{AMC8 box|year=2019|num-b=1|num-a=3}}
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{{MAA Notice}}

Revision as of 01:28, 20 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 people has at least $2$ apples?

Solution 1

Using Stars and bars, and removing $6$ apples so each person can have $2$, we get the total number of ways, which is ${20 \choose 2}$, which is equal to $\boxed{\textbf{(C) }190}$. ~~SmileKat32

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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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