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

(→‎Solution 3)
(→‎Solution 3)
Line 22: Line 22:
  
 
Case <math>1</math>   
 
Case <math>1</math>   
−
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>
+
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
−
<math>a</math> =0
+
<math>a</math> =0,
−
<math>b</math> + <math>c</math> = <math>3</math>
+
<math>b</math> + <math>c</math> = <math>3</math>,
−
<math>b</math> = 0,1,2,3
+
<math>b</math> = 0,1,2,3 ,
 
and respective values of <math>c</math> will be
 
and respective values of <math>c</math> will be
−
<math>c</math> = 3,2,1,0
+
<math>c</math> = 3,2,1,0 ,
 
Which means 4 solutions.
 
Which means 4 solutions.
  
−
Case <math>2</math>   
+
Case <math>2</math>,  
−
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>
+
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
−
<math>a</math> =1
+
<math>a</math> =1,
−
<math>1</math> + <math>b</math> + <math>c</math> = <math>3</math>
+
<math>1</math> + <math>b</math> + <math>c</math> = <math>3</math>,
−
<math>b</math> + <math>c</math> = <math>2</math>
+
<math>b</math> + <math>c</math> = <math>2</math>,
−
<math>b</math> = 0,1,2
+
<math>b</math> = 0,1,2 ,
 
and respective values of <math>c</math> will be
 
and respective values of <math>c</math> will be
−
<math>c</math> = 2,1,0
+
<math>c</math> = 2,1,0 ,
 
Which means 3 solutions.
 
Which means 3 solutions.
  
−
Case <math>3</math>
+
Case <math>3</math> ,
−
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>
+
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
−
<math>a</math>= 2
+
<math>a</math>= 2,
−
<math>2</math> + <math>b</math> + <math>c</math> = <math>3</math>
+
<math>2</math> + <math>b</math> + <math>c</math> = <math>3</math>,
−
<math>b</math> + <math>c</math> = <math>1</math>
+
<math>b</math> + <math>c</math> = <math>1</math>,
−
<math>b</math> = 0,1
+
<math>b</math> = 0,1 ,
 
and respective values of <math>c</math> will be
 
and respective values of <math>c</math> will be
−
<math>c</math> = 1,0
+
<math>c</math> = 1,0 ,
 
Which means 2 solutions.
 
Which means 2 solutions.
  
−
Case <math>4</math>
+
Case <math>4</math>,
−
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>
+
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
−
<math>a</math> = 3
+
<math>a</math> = 3,
−
<math>3</math> + <math>b</math> + <math>c</math> = <math>3</math>
+
<math>3</math> + <math>b</math> + <math>c</math> = <math>3</math>,
−
<math>b</math> + <math>c</math> = <math>0</math>
+
<math>b</math> + <math>c</math> = <math>0</math>,
−
<math>b</math> = 0
+
<math>b</math> = 0 ,
 
and respective value of <math>c</math> will be
 
and respective value of <math>c</math> will be
−
<math>c</math> = 0
+
<math>c</math> = 0 ,
 
Which means 1 solution.
 
Which means 1 solution.
  

Revision as of 11:07, 15 November 2019

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 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 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, 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 casework. Let the three friends be $a$, $b$, $c$ repectively.

Case $1$ $a$ + $b$ + $c$ = $3$, $a$ =0, $b$ + $c$ = $3$, $b$ = 0,1,2,3 , and respective values of $c$ will be $c$ = 3,2,1,0 , Which means 4 solutions.

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

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

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

Therefore there will be total 10 solutions. \boxed{\textbf{(D)}\ 10}.

This is not a fast or an elegant solution but if this comes to your mind in the exam it will be beneficial.

Solution by $phoenixfire$

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

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