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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 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 [[Ball-and-urn]] to find the number of possibilities is <math>\binom{3+3-1}{3} = \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</math>, <math>b</math>, <math>c</math> repectively.
Let the three friends be <math>a, b, c</math> respectively.


<math>a</math> + <math>b</math> + <math>c</math> = <math>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}{3} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}</math>.
<math>\binom{3+2}{2} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.


Solution by <math>phoenixfire</math>
Solution by [[User:phoenixfire|phoenixfire]]
Minor Edit by [[User:Yuvag|Yuvag]]


==Solution 3==
==Solution 3==
Like in solution 1 and solution 2, 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 casework.  
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:
 
<math>\binom{n-1}{s-1}</math>
 
Now we can plug in our values,
 
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 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.  
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.


Case <math>1</math>
<math>a+b +c = 3</math>,
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>
 
<math>a</math> =0
 
<math>b</math> + <math>c</math> = <math>3</math>
Case <math>1:a=0</math>,
<math>b</math> = 0,1,2,3
 
and respective values of <math>c</math> will be
<math>b+c = 3</math>,
<math>c</math> = 3,2,1,0
 
Which means 4 solutions.
<math>b = 0,1,2,3</math> ,
 
<math>c = 3,2,1,0</math>,
 
<math>\boxed{\textbf\ 4}</math> solutions.
 
 
Case <math>2:a=1</math>,
 
<math>1 + b + c = 3</math>,
 
<math>b + c = 2</math>,
 
<math>b = 0,1,2</math> ,
 
<math>c = 2,1,0</math> ,
 
<math>\boxed{\textbf\ 3}</math> solutions.
 
 
Case <math>3:a= 2</math>,
 
<math>2 + b + c = 3</math>,
 
<math>b + c = 1</math>,
 
<math>b = 0,1</math>,
 
<math>c = 1,0</math>,
 
<math>\boxed{\textbf\ 2}</math> solutions.
 
 
Case <math>4:a = 3</math>,
 
<math>3+b+c = 3</math>,


Case <math>2</math> 
<math>b+c = 0</math>,
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>
<math>a</math> =1
<math>1</math> + <math>b</math> + <math>c</math> = <math>3</math>
<math>b</math> + <math>c</math> = <math>2</math>
<math>b</math> = 0,1,2
and respective values of <math>c</math> will be
<math>c</math> = 2,1,0
Which means 3 solutions.


Case <math>3</math>
<math>b = 0</math>,
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>
<math>a</math>= 2
<math>2</math> + <math>b</math> + <math>c</math> = <math>3</math>
<math>b</math> + <math>c</math> = <math>1</math>
<math>b</math> = 0,1
and respective values of <math>c</math> will be
<math>c</math> = 1,0
Which means 2 solutions.


Case <math>4</math>
<math>c = 0</math>,
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>
<math>a</math> = 3
<math>3</math> + <math>b</math> + <math>c</math> = <math>3</math>
<math>b</math> + <math>c</math> = <math>0</math>
<math>b</math> = 0
and respective value of <math>c</math> will be
<math>c</math> = 0
Which means 1 solution.


Therefore there will be total 10 solutions. \boxed{\textbf{(D)}\ 10}.
<math>\boxed{\textbf\ 1}</math> solution.


This is not a fast or an elegant solution but if this comes to your mind in the exam it will be beneficial.
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 <math>phoenixfire</math>
==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: File missing

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