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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>.
+
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}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.
 +
 
 +
Solution by [[User:phoenixfire|phoenixfire]]
 +
Minor Edits by [[User:Yuvag|Yuvag]] : "... is <math>\binom{3+3-1}{3} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}</math>." to "... is <math>\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>."
 +
 
 +
All credit still goes to phoenixfire.
  
 
==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 pencils 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-negative integral solutions.  
−
Let the three friends be <math>a</math>, <math>b</math>, <math>c</math> repectively.
+
Let the three friends be <math>a, b, c</math> respectively.
  
−
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>
+
<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}{2} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.
 +
 
 +
Solution by [[User:phoenixfire|phoenixfire]]
  
−
Solution by <math>phoenixfire</math>
+
Minor Edits by [[User:Yuvag|Yuvag]] : "<math>\binom{3+2}{3} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}</math>." to "<math>\binom{3+2}{2} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.
 +
 
 +
All credit still goes to 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.  
+
For each of the 3 People to have at least one pencil when distributing 6 pencil amongst them, we can use another formula from the [[Ball-and-urn]] counting technique, shown below:
 +
 
 +
 
 +
 
 +
for n = number of items, and s = slots:
 +
 
 +
 
 +
<math>\binom{n-1}{s-1}</math>
 +
 
 +
 
 +
Now we can plug in our values,
 +
 
 +
number of items = 6, and slots = 3:
 +
 
 +
 
 +
<math>\binom{6-1}{3-1}    =    \binom{5}{2}    =    \boxed{\textbf{(D)}\ 10}</math>.
 +
 
 +
 
 +
 
 +
 
 +
Solution by [[User:Yuvag|Yuvag]]
 +
 
 +
==Solution 4==
 +
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>
+
<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>
+
Case <math>1:a=0</math>,
−
<math>b</math> = 0,1,2,3
+
 
−
and respective values of <math>c</math> will be
+
<math>b + c = 3</math>,
−
<math>c</math> = 3,2,1,0
+
 
−
Which means 4 solutions.
+
<math>b = 0,1,2,3</math> ,
 +
 
 +
<math>c = 3,2,1,0</math>,
 +
 
 +
<math>\boxed{\textbf\ 4}</math> solutions.
 +
 
 +
 
 +
Case <math>2:a=1</math>,
 +
 
 +
<math>1 + b + c = 3</math>,
 +
 
 +
<math>b + c = 2</math>,
 +
 
 +
<math>b = 0,1,2</math> ,
 +
 
 +
<math>c = 2,1,0</math> ,
 +
 
 +
<math>\boxed{\textbf\ 3}</math> solutions.
 +
 
 +
 
 +
Case <math>3:a= 2</math>,
 +
 
 +
<math>2 + b + c = 3</math>,
 +
 
 +
<math>b + c = 1</math>,
 +
 
 +
<math>b = 0,1</math>,
 +
 
 +
<math>c = 1,0</math>,
 +
 
 +
<math>\boxed{\textbf\ 2}</math> solutions.
 +
 
 +
 
 +
Case <math>4:a = 3</math>,
 +
 
 +
<math>3 + b + c = 3</math>,
 +
 
 +
<math>b + c = 0</math>,
 +
 
 +
<math>b = 0</math>,
  
−
Case <math>2</math> 
+
<math>c = 0</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.
 
  
−
Case <math>3</math>
+
<math>\boxed{\textbf\ 1}</math> solution.
−
<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.
 
  
−
Case <math>4</math>
+
Therefore there will be a total  of <math>4+3+2+1=10</math> solutions. <math>\boxed{\textbf{(D)}\ 10}</math>.
−
<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>
+
Solution by [[User:phoenixfire|phoenixfire]]
−
<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}.
+
==Video Solution==
 +
https://youtu.be/FUnwTLP7gr0  Soo, DRMS, NM
  
−
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 <math>phoenixfire</math>
 
  
 
==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}}

Latest revision as of 23:35, 29 December 2023

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 $\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}$.

Solution by phoenixfire 

Minor Edits by Yuvag : "... is $\binom{3+3-1}{3} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}$." to "... is $\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}$."

All credit still goes to phoenixfire.

Solution 2

Like in solution 1, for each person to have at least one pencil, assign one of the pencils 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-negative integral solutions. Let the three friends be $a, b, c$ respectively.

$a + b + c = 3$ The total being 3 and 2 plus signs, which implies $\binom{3+2}{2} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}$.

Solution by phoenixfire

Minor Edits by Yuvag : "$\binom{3+2}{3} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}$." to "$\binom{3+2}{2} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}$.

All credit still goes to phoenixfire.

Solution 3

For each of the 3 People to have at least one pencil when distributing 6 pencil amongst them, we can use another formula from the Ball-and-urn counting technique, shown below:


for n = number of items, and s = slots: 


$\binom{n-1}{s-1}$


Now we can plug in our values,

number of items = 6, and slots = 3:


$\binom{6-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}$.



Solution by Yuvag

Solution 4

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 $4+3+2+1=10$ solutions. $\boxed{\textbf{(D)}\ 10}$. Solution by phoenixfire

Video Solution

https://youtu.be/FUnwTLP7gr0 Soo, DRMS, NM


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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