Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2004 AMC 8 Problems/Problem 17"
m (→‎Solution 2)
 
(One intermediate revision by one other user not shown)
Line 13: Line 13:
  
 
==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, b, c</math> repectively.
+
Let the three friends be <math>a, b, c</math> respectively.
  
 
<math>a + b + c = 3</math>
 
<math>a + b + c = 3</math>
Line 105: Line 105:
 
<math>\boxed{\textbf\ 1}</math> solution.
 
<math>\boxed{\textbf\ 1}</math> solution.
  
−
Therefore there will be a total  of 10 solutions. <math>\boxed{\textbf{(D)}\ 10}</math>.
+
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]]
 
Solution by [[User:phoenixfire|phoenixfire]]
−
 
  
 
==Video Solution==
 
==Video Solution==

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_8_Problems/Problem_17&oldid=208844"