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

(→‎Solution 3)
(→‎Solution 3)
Line 23: Line 23:
 
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 <math>\boxed{\textbf\4}</math> 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 <math>\boxed{\textbf\3}</math> 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 <math>\boxed{\textbf\2}</math> 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.
 
  
−
Therefore there will be total 10 solutions. \boxed{\textbf{(D)}\ 10}.
+
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.
 
This is not a fast or an elegant solution but if this comes to your mind in the exam it will be beneficial.

Revision as of 11:11, 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 $\boxed{\textbf\4}$ (Error compiling LaTeX. Unknown error_msg) 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 $\boxed{\textbf\3}$ (Error compiling LaTeX. Unknown error_msg) 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 $\boxed{\textbf\2}$ (Error compiling LaTeX. Unknown error_msg) 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 $\boxed{\textbf\1}$ (Error compiling LaTeX. Unknown error_msg) 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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