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

Revision as of 10:59, 4 January 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 $\binom{3+3-1}{3-1} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}$ .

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

Solution 3

For each person to have at least one pencil, we can use another formula from the Ball-and-urn counting technique:

for $\binom{$ (Error compiling LaTeX. Unknown error_msg)n =$number of items}{$ (Error compiling LaTeX. Unknown error_msg)s =$slots}$ (Error compiling LaTeX. Unknown error_msg)

$\binom{6-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}$ .

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

@@ Line 22: / Line 22: @@
 For each person to have at least one pencil, we can use another formula from the [[Ball-and-urn]] counting technique:
-for <math>\binom{n = number of items}{ s = slots}</math>
-<math>\binom{6-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>.
+for <math>\binom{</math>n =<math> number of items}{</math>s =<math> slots}</math>
+<math>\binom{6-1}{3-1}     =     \binom{5}{2}     =    \boxed{\textbf{(D)}\ 10}</math>.

2004 AMC 8 (Problems • Answer Key • Resources)
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