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

Revision as of 17:39, 5 January 2024

1 Problem 25
2 Solution 1 (Stars and Bars/Sticks and Stones)
3 Solution 2 (Answer Choices)
4 Solution 3
5 Video Solution by Math-X (Let's review stars and bars then do this under a 1 minute!!!)
6 Video Solution by OmegaLearn
7 Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
8 Video Solutions
9 See also

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? $\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$

Solution 1 (Stars and Bars/Sticks and Stones)

This method uses the counting method of stars and bars. Since each person must have at least $2$ apples, we can remove $2*3$ apples from the total that need to be sorted. With the remaining $18$ apples, we can use stars and bars to determine the number of possibilities. Assume there are $18$ stars in a row, and $2$ bars, which will be placed to separate the stars into groups of $3$ . In total, there are $18$ spaces for stars $+ 2$ spaces for bars, for a total of $20$ spaces. We can now do $20 \choose 2$ . This is because if we choose distinct $2$ spots for the bars to be placed, each combo of $3$ groups will be different, and all apples will add up to $18$ . We can also do this because the apples are indistinguishable. $20 \choose 2$ is $190$ , therefore the answer is $\boxed{\textbf{(C) }190}$ .

~goofytaipan91

Solution 2 (Answer Choices)

Consider an unordered triple $(a,b,c)$ where $a+b+c=24$ and $a,b,c$ are not necessarily distinct. Then, we will either have $1$ , $3$ , or $6$ distinguishable ways to assign $a$ , $b$ , and $c$ to Alice, Becky, and Chris. Thus, our answer will be $x+3y+6z$ for some nonnegative integers $x,y,z$ . Notice that we only have $1$ way to assign the numbers $a,b,c$ to Alice, Becky, and Chris when $a=b=c$ . As this only happens $1$ way ( $a=b=c=8$ ), our answer is $1+3y+6z$ for some $y,z$ . Finally, notice that this implies the answer is $1$ mod $3$ . The only answer choice that satisfies this is $\boxed{\textbf{(C) }190}$ .

-BorealBear

Solution 3

Since each person needs to have at least two apples, we can simply give each person two, leaving $24 - 2\times3=18$ apples. For the remaining apples, if Alice is going to have $a$ apples, Becky is going to have $b$ apples, and Chris is going to have $c$ apples, we have indeterminate equation $a+b+c=18$ . Currently, we can see that $0 \leq a\leq 18$ where $a$ is an integer, and when $a$ equals any number in the range, there will be $18-a+1=19-a$ sets of values for $b$ and $c$ . Thus, there are $19 + 18 + 17 + \cdots + 1 = \boxed{\textbf{(C) }190}$ possible sets of values in total.

~Bloggish

Video Solution by Math-X (Let's review stars and bars then do this under a 1 minute!!!)

https://youtu.be/IgpayYB48C4?si=FbCF4qyqg6UtEljw&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(Problem 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

@@ Line 2: / Line 2: @@
 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 1 (Stars and Bars/Sticks and Stones)==
+This method uses the counting method of stars and bars. 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)==

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
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