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

(→‎Solution 2)
(→‎Solution 2)
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==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.
 
Let the three friends be <math>a</math>, <math>b</math>, <math>c</math> repectively.
  
Line 16: Line 16:
  
 
Solution by <math>phoenixfire</math>
 
Solution by <math>phoenixfire</math>
 +
 +
==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.
 +
Let the three friends be <math>a</math>, <math>b</math>, <math>c</math> repectively.
 +
 +
<math>Case 1</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 4 solutions.
 +
 +
<math>Case 2</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.
 +
 +
<math>Case 3</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.
 +
 +
<math>Case 4</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}$.
 +
 +
This is not a fast or an elegant solution but if this comes to your mind in the exam it will be beneficial.
  
 
==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 11:02, 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.

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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