Difference between revisions of "2004 AMC 8 Problems/Problem 17"

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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>.
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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} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}</math>.
  
 
==Solution 2==
 
==Solution 2==
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==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.  
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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> 
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<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>,
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Case <math>1:a=0</math>,
  
<math>b</math> = 0,1,2,3 ,
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<math>b + c = 3</math>,
  
and respective values of <math>c</math> will be
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<math>b = 0,1,2,3</math> ,
<math>c</math> = 3,2,1,0 ,
 
  
Which means <math>\boxed{\textbf\ 4}</math> solutions.
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<math>c = 3,2,1,0</math>,
  
Case <math>2</math>
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<math>\boxed{\textbf\ 4}</math> solutions.
  
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
 
  
<math>a</math> =1,
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Case <math>2:a=1</math>,
<math>1</math> + <math>b</math> + <math>c</math> = <math>3</math>,
 
  
<math>b</math> + <math>c</math> = <math>2</math>,
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<math>1 + b + c = 3</math>,
  
<math>b</math> = 0,1,2 ,
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<math>b + c = 2</math>,
  
and respective values of <math>c</math> will be
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<math>b = 0,1,2</math> ,
<math>c</math> = 2,1,0 ,
 
  
Which means <math>\boxed{\textbf\ 3}</math> solutions.
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<math>c = 2,1,0</math> ,
  
Case <math>3</math> ,
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<math>\boxed{\textbf\ 3}</math> solutions.
  
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
 
  
<math>a</math>= 2,
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Case <math>3:a= 2</math>,
  
<math>2</math> + <math>b</math> + <math>c</math> = <math>3</math>,
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<math>2 + b + c = 3</math>,
  
<math>b</math> + <math>c</math> = <math>1</math>,
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<math>b + c = 1</math>,
  
<math>b</math> = 0,1 ,
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<math>b = 0,1</math>,
  
and respective values of <math>c</math> will be
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<math>c = 1,0</math>,
<math>c</math> = 1,0 ,
 
  
Which means <math>\boxed{\textbf\ 2}</math> solutions.
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<math>\boxed{\textbf\ 2}</math> solutions.
  
Case <math>4</math>,
 
  
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
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Case <math>4:a = 3</math>,
  
<math>a</math> = 3,
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<math>3 + b + c = 3</math>,
  
<math>3</math> + <math>b</math> + <math>c</math> = <math>3</math>,
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<math>b + c = 0</math>,
  
<math>b</math> + <math>c</math> = <math>0</math>,
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<math>b = 0</math>,
  
<math>b</math> = 0 ,
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<math>c = 0</math>,
  
and respective value of <math>c</math> will be
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<math>\boxed{\textbf\ 1}</math> solution.
<math>c</math> = 0 ,
 
 
 
Which means <math>\boxed{\textbf\ 1}</math> solution.
 
 
 
Therefore there will be total 10 solutions. <math>\boxed{\textbf{(D)}\ 10}</math>.
 
 
 
This is not a fast or an elegant solution but if this comes to your mind in the exam it will be beneficial.
 
  
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Therefore there will be a total  of 10 solutions. <math>\boxed{\textbf{(D)}\ 10}</math>.
 
Solution by [[User:phoenixfire|phoenixfire]]
 
Solution by [[User:phoenixfire|phoenixfire]]
  

Revision as of 14:25, 6 July 2021

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

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 $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use number of non-negetive integral soutions. Let the three friends be $a, b, c$ repectively.

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

Solution by phoenixfire

Solution 3

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 10 solutions. $\boxed{\textbf{(D)}\ 10}$. Solution by phoenixfire

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
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

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