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2004 AMC 8 Problems/Problem 17: Difference between revisions

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==Solution 1==
==Solution 1==
For each person to have at least one pencil, assign one pencil to each of the three friends so that you have <math>3</math> left. In partitioning the remaining <math>3</math> pencils into <math>3</math> distinct groups, use [[Ball-and-urn]] to find the number of possibilities is <math>\binom{3+3-1}{3-1} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}</math>.
For each person to have at least one pencil, assign one pencil to each of the three friends so that you have <math>3</math> left. In partitioning the remaining <math>3</math> pencils into <math>3</math> distinct groups, use [[Ball-and-urn]] to find the number of possibilities is <math>\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.
 
Solution by [[User:phoenixfire|phoenixfire]]
Minor Edit by [[User:Yuvag|Yuvag]]


==Solution 2==
==Solution 2==
Like in solution 1, for each person to have at least one pencil, assign one of the pencil to each of the three friends so that you have <math>3</math> left. In partitioning the remaining <math>3</math> pencils into <math>3</math> distinct groups, use number of non-negetive integral soutions.  
Like in solution 1, for each person to have at least one pencil, assign one of the pencils to each of the three friends so that you have <math>3</math> left. In partitioning the remaining <math>3</math> pencils into <math>3</math> distinct groups, use number of non-negative integral solutions.  
Let the three friends be <math>a, b, c</math> repectively.
Let the three friends be <math>a, b, c</math> respectively.


<math>a + b + c = 3</math>
<math>a+b+c = 3</math>
The total being 3 and 2 plus signs, which implies
The total being 3 and 2 plus signs, which implies
<math>\binom{3+2}{2} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.
<math>\binom{3+2}{2} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.


Solution by [[User:phoenixfire|phoenixfire]]
Solution by [[User:phoenixfire|phoenixfire]]
Minor Edit by [[User:Yuvag|Yuvag]]


==Solution 3==
For each of the 3 People to have at least one pencil when distributing 6 pencil amongst them, we can use another formula from the [[Ball-and-urn]] counting technique, shown below:


for n = number of items, and s = slots:


==Solution 3==
<math>\binom{n-1}{s-1}</math>
For each person to have at least one pencil, we can use another formula from the [[Ball-and-urn]] counting technique:


for <math>\binom{n = number of items}{ s = slots}</math>
Now we can plug in our values,


<math>\binom{6-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.
number of items = 6, and slots = 3:


<math>\binom{6-1}{3-1}    =    \binom{5}{2}    =    \boxed{\textbf{(D)}\ 10}</math>.


Solution by [[User:Yuvag|Yuvag]]
Solution by [[User:Yuvag|Yuvag]]


==Solution 3==
==Solution 4==
Like in solution 1 and solution 2, assign one pencil to each of the three friends so that you have <math>3</math> left. In partitioning the remaining <math>3</math> pencils into <math>3</math> distinct groups use casework.  
Like in solution 1 and solution 2, assign one pencil to each of the three friends so that you have <math>3</math> left. In partitioning the remaining <math>3</math> pencils into <math>3</math> distinct groups use casework.  
Let the three friends be <math>a</math>, <math>b</math>, <math>c</math> repectively.
Let the three friends be <math>a</math>, <math>b</math>, <math>c</math> repectively.


<math>a + b + c = 3</math>,
<math>a+b +c = 3</math>,




Case <math>1:a=0</math>,
Case <math>1:a=0</math>,


<math>b + c = 3</math>,
<math>b+c = 3</math>,


<math>b = 0,1,2,3</math> ,
<math>b = 0,1,2,3</math> ,
Line 75: Line 82:
Case <math>4:a = 3</math>,
Case <math>4:a = 3</math>,


<math>3 + b + c = 3</math>,
<math>3+b+c = 3</math>,


<math>b + c = 0</math>,
<math>b+c = 0</math>,


<math>b = 0</math>,
<math>b = 0</math>,
Line 85: Line 92:
<math>\boxed{\textbf\ 1}</math> solution.
<math>\boxed{\textbf\ 1}</math> solution.


Therefore there will be a total of 10 solutions. <math>\boxed{\textbf{(D)}\ 10}</math>.
Therefore there will be a total of <math>4+3+2+1=\boxed{\textbf{(D)}\ 10}</math> solutions.
Solution by [[User:phoenixfire|phoenixfire]]
Solution by [[User:phoenixfire|phoenixfire]]
==Video Solution==
https://youtu.be/FUnwTLP7gr0  Soo, DRMS, NM


==See Also==
==See Also==
{{AMC8 box|year=2004|num-b=16|num-a=18}}
{{AMC8 box|year=2004|num-b=16|num-a=18}}
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 19:46, 29 December 2024

Problem

Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?

$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

Solution 1

For each person to have at least one pencil, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use Ball-and-urn to find the number of possibilities is $\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}$.

Solution by phoenixfire Minor Edit by Yuvag

Solution 2

Like in solution 1, for each person to have at least one pencil, assign one of the pencils to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use number of non-negative integral solutions. Let the three friends be $a, b, c$ respectively.

$a+b+c = 3$ The total being 3 and 2 plus signs, which implies $\binom{3+2}{2} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}$.

Solution by phoenixfire Minor Edit by Yuvag

Solution 3

For each of the 3 People to have at least one pencil when distributing 6 pencil amongst them, we can use another formula from the Ball-and-urn counting technique, shown below:

for n = number of items, and s = slots: 

$\binom{n-1}{s-1}$

Now we can plug in our values,

number of items = 6, and slots = 3:

$\binom{6-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}$.

Solution by Yuvag

Solution 4

Like in solution 1 and solution 2, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups use casework. Let the three friends be $a$, $b$, $c$ repectively.

$a+b +c = 3$,


Case $1:a=0$,

$b+c = 3$,

$b = 0,1,2,3$ ,

$c = 3,2,1,0$,

$\boxed{\textbf\ 4}$ solutions.


Case $2:a=1$,

$1 + b + c = 3$,

$b + c = 2$,

$b = 0,1,2$ ,

$c = 2,1,0$ ,

$\boxed{\textbf\ 3}$ solutions.


Case $3:a= 2$,

$2 + b + c = 3$,

$b + c = 1$,

$b = 0,1$,

$c = 1,0$,

$\boxed{\textbf\ 2}$ solutions.


Case $4:a = 3$,

$3+b+c = 3$,

$b+c = 0$,

$b = 0$,

$c = 0$,

$\boxed{\textbf\ 1}$ solution.

Therefore there will be a total of $4+3+2+1=\boxed{\textbf{(D)}\ 10}$ solutions. Solution by phoenixfire

Video Solution

https://youtu.be/FUnwTLP7gr0 Soo, DRMS, NM

See Also

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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