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

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 ." to "... is ."

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:

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

@@ Line 4: / Line 4: @@
 <math>\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12</math>
-==Solution==
+==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>_{3+3-1} C _{3-1} = _10 C _2 = \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==
+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, b, c</math> respectively.
+<math>a + b + c = 3</math>
+The total being 3 and 2 plus signs, which implies
+<math>\binom{3+2}{2} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.
+Solution by [[User:phoenixfire|phoenixfire]]
+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==
+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.
+<math>a + b + c = 3</math>,
+Case <math>1:a=0</math>,
+<math>b + c = 3</math>,
+<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>,
+<math>c = 0</math>,
+<math>\boxed{\textbf\ 1}</math> solution.
+Therefore there will be a total  of <math>4+3+2+1=10</math> solutions. <math>\boxed{\textbf{(D)}\ 10}</math>.
+Solution by [[User:phoenixfire|phoenixfire]]
+==Video Solution==
+https://youtu.be/FUnwTLP7gr0  Soo, DRMS, NM
 ==See Also==
 {{AMC8 box|year=2004|num-b=16|num-a=18}}
 {{MAA Notice}}

2004 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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