# 2004 AMC 8 Problems/Problem 17

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## 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 pencil 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-negetive integral soutions. Let the three friends be $a, b, c$ repectively. $a + b + c = 3$ 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 10 solutions. $\boxed{\textbf{(D)}\ 10}$. Solution by phoenixfire

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