2004 AMC 8 Problems/Problem 17: Difference between revisions

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 13: / Line 13: @@
 ==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>

2004 AMC 8 (Problems • Answer Key • Resources)
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