Difference between revisions of "2019 AMC 8 Problems/Problem 25"
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| − | ==Problem | + | ==Problem== |
Alice has <math>24</math> 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? | Alice has <math>24</math> 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? | ||
| − | + | <math>\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380</math> | |
| − | |||
| − | ==Solution | + | ==Solution 1 (Stars and Bars/Sticks and Stones)== |
| − | |||
| − | + | Note: This solution uses the non-negative version for stars and bars. A solution using the positive version of stars is similar (first removing an apple from each person instead of 2). | |
| − | + | This method uses the counting method of stars and bars (non-negative version). Since each person must have at least <math>2</math> apples, we can remove <math>2*3</math> apples from the total that need to be sorted. With the remaining <math>18</math> apples, we can use stars and bars to determine the number of possibilities. Assume there are <math>18</math> stars in a row, and <math>2</math> bars, which will be placed to separate the stars into groups of <math>3</math>. In total, there are <math>18</math> spaces for stars <math>+ 2</math> spaces for bars, for a total of <math>20</math> spaces. We can now do <math>20 \choose 2</math>. This is because if we choose distinct <math>2</math> spots for the bars to be placed, each combo of <math>3</math> groups will be different, and all apples will add up to <math>18</math>. We can also do this because the apples are indistinguishable. <math>20 \choose 2</math> is <math>190</math>, therefore the answer is <math>\boxed{\textbf{(C) }190}</math>. | |
| − | {{ | + | |
| + | ~goofytaipan91 | ||
| + | |||
| + | ==Solution 2 (Answer Choices)== | ||
| + | Consider an unordered triple <math> (a,b,c) </math> where <math> a+b+c=24 </math> and <math> a,b,c </math> are not necessarily distinct. Then, we will either have <math> 1 </math>, <math> 3 </math>, or <math> 6 </math> distinguishable ways to assign <math> a </math>, <math> b </math>, and <math> c </math> to Alice, Becky, and Chris. Thus, our answer will be <math> x+3y+6z </math> for some nonnegative integers <math> x,y,z </math>. Notice that we only have <math> 1 </math> way to assign the numbers <math> a,b,c </math> to Alice, Becky, and Chris when <math> a=b=c </math>. As this only happens <math> 1 </math> way (<math>a=b=c=8</math>), our answer is <math> 1+3y+6z </math> for some <math> y,z </math>. Finally, notice that this implies the answer is <math> 1 </math> mod <math> 3 </math>. The only answer choice that satisfies this is <math> \boxed{\textbf{(C) }190} </math>. | ||
| + | |||
| + | -BorealBear | ||
| + | |||
| + | ==Solution 3== | ||
| + | Since each person needs to have at least two apples, we can simply give each person two, leaving <math> 24 - 2\times3=18 </math> apples. For the remaining apples, if Alice is going to have <math> a </math> apples, Becky is going to have <math> b </math> apples, and Chris is going to have <math> c </math> apples, we have indeterminate equation <math> a+b+c=18 </math>. Currently, we can see that <math> 0 \leq a\leq 18 </math> where <math> a </math> is an integer, and when <math> a </math> equals any number in the range, there will be <math> 18-a+1=19-a </math> sets of values for <math> b </math> and <math> c </math>. Thus, there are <math> 19 + 18 + 17 + \cdots + 1 = \boxed{\textbf{(C) }190} </math> possible sets of values in total. | ||
| + | |||
| + | ~[[User:Bloggish|Bloggish]] | ||
| + | |||
| + | ==Video Solution by Math-X (Let's review stars and bars together first!!!)== | ||
| + | https://youtu.be/IgpayYB48C4?si=SzBgzW4jHelkYwP1&t=8105 | ||
| + | |||
| + | ~Math-X | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/5UojVH4Cqqs?t=5131 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==Video Solution by The Power of Logic(1 to 25 Full Solution)== | ||
| + | https://youtu.be/Xm4ZGND9WoY | ||
| + | |||
| + | ~Hayabusa1 | ||
| + | |||
| + | ==Video Solutions== | ||
| + | |||
| + | https://www.youtube.com/watch?v=EJzSOPXULBc | ||
| + | |||
| + | - Happytwin | ||
| + | |||
| + | https://www.youtube.com/watch?v=wJ7uvypbB28 | ||
| + | |||
| + | https://www.youtube.com/watch?v=2dBUklyUaNI | ||
| + | |||
| + | |||
| + | https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7 | ||
| + | |||
| + | ~ MathEx | ||
| + | |||
| + | https://youtu.be/8kzjB60pBrA | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC8 box|year=2019|num-b=24|after=Last Problem}} | ||
| + | [[Category:Introductory Combinatorics Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 09:33, 9 November 2024
Contents
Problem
Alice has 
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 (Stars and Bars/Sticks and Stones)
Note: This solution uses the non-negative version for stars and bars. A solution using the positive version of stars is similar (first removing an apple from each person instead of 2).
This method uses the counting method of stars and bars (non-negative version). Since each person must have at least 
apples, we can remove 
apples from the total that need to be sorted. With the remaining 
apples, we can use stars and bars to determine the number of possibilities. Assume there are stars in a row, and bars, which will be placed to separate the stars into groups of 
. In total, there are spaces for stars 
spaces for bars, for a total of 
spaces. We can now do 
. This is because if we choose distinct spots for the bars to be placed, each combo of groups will be different, and all apples will add up to . We can also do this because the apples are indistinguishable. is 
, therefore the answer is 
.
~goofytaipan91
Solution 2 (Answer Choices)
Consider an unordered triple 
where 
and 
are not necessarily distinct. Then, we will either have 
, , or 
distinguishable ways to assign 
, 
, and 
to Alice, Becky, and Chris. Thus, our answer will be 
for some nonnegative integers 
. Notice that we only have way to assign the numbers to Alice, Becky, and Chris when 
. As this only happens way (
), our answer is 
for some 
. Finally, notice that this implies the answer is mod . The only answer choice that satisfies this is .
-BorealBear
Solution 3
Since each person needs to have at least two apples, we can simply give each person two, leaving 
apples. For the remaining apples, if Alice is going to have apples, Becky is going to have apples, and Chris is going to have apples, we have indeterminate equation 
. Currently, we can see that 
where is an integer, and when equals any number in the range, there will be 
sets of values for and . Thus, there are 
possible sets of values in total.
Video Solution by Math-X (Let's review stars and bars together first!!!)
https://youtu.be/IgpayYB48C4?si=SzBgzW4jHelkYwP1&t=8105
~Math-X
Video Solution by OmegaLearn
https://youtu.be/5UojVH4Cqqs?t=5131
~ pi_is_3.14
Video Solution by The Power of Logic(1 to 25 Full Solution)
~Hayabusa1
Video Solutions
https://www.youtube.com/watch?v=EJzSOPXULBc
- Happytwin
https://www.youtube.com/watch?v=wJ7uvypbB28
https://www.youtube.com/watch?v=2dBUklyUaNI
https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7
~ MathEx
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
