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==
 
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 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.  
Let the three friends be <math>a</math>, <math>b</math>, <math>c</math> repectively.
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Let the three friends be <math>a, b, c</math> repectively.
  
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>
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<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}{3} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}</math>.
  
Solution by <math>phoenixfire</math>
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Solution by [[User:phoenixfire|phoenixfire]]
  
 
==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.  
+
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>,
 
 
 
<math>b</math> = 0,1,2,3 ,
 
  
and respective values of <math>c</math> will be
 
<math>c</math> = 3,2,1,0 ,
 
  
Which means <math>\boxed{\textbf\ 4}</math> solutions.
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Case <math>1:a=0</math>,
  
Case <math>2</math>,
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<math>b + c = 3</math>,
  
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
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<math>b = 0,1,2,3</math> ,
  
<math>a</math> =1,
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<math>c = 3,2,1,0</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>\boxed{\textbf\ 4}</math> solutions.
  
<math>b</math> = 0,1,2 ,
 
  
and respective values of <math>c</math> will be
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Case <math>2:a=1</math>,
<math>c</math> = 2,1,0 ,
 
  
Which means <math>\boxed{\textbf\ 3}</math> solutions.
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<math>1 + b + c = 3</math>,
  
Case <math>3</math> ,
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<math>b + c = 2</math>,
  
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
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<math>b = 0,1,2</math> ,
  
<math>a</math>= 2,
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<math>c = 2,1,0</math> ,
  
<math>2</math> + <math>b</math> + <math>c</math> = <math>3</math>,
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<math>\boxed{\textbf\ 3}</math> solutions.
  
<math>b</math> + <math>c</math> = <math>1</math>,
 
  
<math>b</math> = 0,1 ,
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Case <math>3:a= 2</math>,
  
and respective values of <math>c</math> will be
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<math>2 + b + c = 3</math>,
<math>c</math> = 1,0 ,
 
  
Which means <math>\boxed{\textbf\ 2}</math> solutions.
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<math>b + c = 1</math>,
  
Case <math>4</math>,
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<math>b = 0,1</math>,
  
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
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<math>c = 1,0</math>,
  
<math>a</math> = 3,
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<math>\boxed{\textbf\ 2}</math> solutions.
  
<math>3</math> + <math>b</math> + <math>c</math> = <math>3</math>,
 
  
<math>b</math> + <math>c</math> = <math>0</math>,
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Case <math>4:a = 3</math>,
  
<math>b</math> = 0 ,
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<math>3 + b + c = 3</math>,
  
and respective value of <math>c</math> will be
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<math>b + c = 0</math>,
<math>c</math> = 0 ,
 
  
Which means <math>\boxed{\textbf\ 1}</math> solution.
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<math>b = 0</math>,
  
Therefore there will be total 10 solutions. <math>\boxed{\textbf{(D)}\ 10}</math>.
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<math>c = 0</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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<math>\boxed{\textbf\ 1}</math> solution.
  
Solution by [[User:a1b2|a1b2]]
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Therefore there will be a total  of 10 solutions. <math>\boxed{\textbf{(D)}\ 10}</math>.
 +
Solution by [[User:phoenixfire|phoenixfire]]
  
 
==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}}

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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