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2004 AMC 8 Problems/Problem 17: Difference between revisions

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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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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