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

m (→‎Solution 2)
m (Formatting)
 
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Solution by [[User:phoenixfire|phoenixfire]]
 
Solution by [[User:phoenixfire|phoenixfire]]
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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>."
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Minor Edit by [[User:Yuvag|Yuvag]]
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All credit still goes to phoenixfire.
 
  
 
==Solution 2==
 
==Solution 2==
Line 16: Line 14:
 
Let the three friends be <math>a, b, c</math> respectively.
 
Let the three friends be <math>a, b, c</math> respectively.
  
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<math>a + b + c = 3</math>
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<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}{2} = \binom{5}{2} = \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 [[User:phoenixfire|phoenixfire]]
−
 
+
Minor Edit by [[User:Yuvag|Yuvag]]
−
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==
 
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 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:
 
for n = number of items, and s = slots:
−
 
  
 
  <math>\binom{n-1}{s-1}</math>
 
  <math>\binom{n-1}{s-1}</math>
−
 
  
 
Now we can plug in our values,  
 
Now we can plug in our values,  
  
 
number of items = 6, and slots = 3:
 
number of items = 6, and slots = 3:
−
 
  
 
<math>\binom{6-1}{3-1}    =    \binom{5}{2}    =    \boxed{\textbf{(D)}\ 10}</math>.
 
<math>\binom{6-1}{3-1}    =    \binom{5}{2}    =    \boxed{\textbf{(D)}\ 10}</math>.
−
 
−
 
−
 
  
 
Solution by [[User:Yuvag|Yuvag]]
 
Solution by [[User:Yuvag|Yuvag]]
Line 53: Line 40:
 
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.
  
−
<math>a + b + c = 3</math>,
+
<math>a+b +c = 3</math>,
  
  
 
Case <math>1:a=0</math>,
 
Case <math>1:a=0</math>,
  
−
<math>b + c = 3</math>,
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<math>b+c = 3</math>,
  
 
<math>b = 0,1,2,3</math> ,
 
<math>b = 0,1,2,3</math> ,
Line 95: Line 82:
 
Case <math>4:a = 3</math>,
 
Case <math>4:a = 3</math>,
  
−
<math>3 + b + c = 3</math>,
+
<math>3+b+c = 3</math>,
  
−
<math>b + c = 0</math>,
+
<math>b+c = 0</math>,
  
 
<math>b = 0</math>,
 
<math>b = 0</math>,
Line 105: Line 92:
 
<math>\boxed{\textbf\ 1}</math> solution.
 
<math>\boxed{\textbf\ 1}</math> solution.
  
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Therefore there will be a total of <math>4+3+2+1=10</math> solutions. <math>\boxed{\textbf{(D)}\ 10}</math>.
+
Therefore there will be a total of <math>4+3+2+1=\boxed{\textbf{(D)}\ 10}</math> solutions.
 
Solution by [[User:phoenixfire|phoenixfire]]
 
Solution by [[User:phoenixfire|phoenixfire]]
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/FUnwTLP7gr0  Soo, DRMS, NM
 
https://youtu.be/FUnwTLP7gr0  Soo, DRMS, NM
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−
 
  
 
==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 19:46, 29 December 2024

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 Edit by Yuvag

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 Edit by Yuvag

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