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

Revision as of 01:29, 12 August 2020

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 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 Ball-and-urn to find the number of possibilities is $\binom{3+3-1}{3} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}$ .

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

Solution by phoenixfire

Solution 3

Like in solution 1 and solution 2, 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 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 21: / Line 21: @@
 Let the three friends be <math>a</math>, <math>b</math>, <math>c</math> repectively.
-Case <math>1</math>
+<math>a + b + c = 3</math>,
-<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
-<math>a</math> =0,
-<math>b</math> + <math>c</math> = <math>3</math>,
+Case <math>1:a=0</math>,
-<math>b</math> = 0,1,2,3 ,
+<math>b + c = 3</math>,
-and respective values of <math>c</math> will be
+<math>b = 0,1,2,3</math> ,
-<math>c</math> = 3,2,1,0 ,
-Which means <math>\boxed{\textbf\ 4}</math> solutions.
+<math>c = 3,2,1,0</math>,
-Case <math>2</math>,
+<math>\boxed{\textbf\ 4}</math> solutions.
-<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
-<math>a</math> =1,
+Case <math>2:a=1</math>,
-<math>1</math> + <math>b</math> + <math>c</math> = <math>3</math>,
-<math>b</math> + <math>c</math> = <math>2</math>,
+<math>1 + b + c = 3</math>,
-<math>b</math> = 0,1,2 ,
+<math>b + c = 2</math>,
-and respective values of <math>c</math> will be
+<math>b = 0,1,2</math> ,
-<math>c</math> = 2,1,0 ,
-Which means <math>\boxed{\textbf\ 3}</math> solutions.
+<math>c = 2,1,0</math> ,
-Case <math>3</math> ,
+<math>\boxed{\textbf\ 3}</math> solutions.
-<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
-<math>a</math>= 2,
+Case <math>3:a= 2</math>,
-<math>2</math> + <math>b</math> + <math>c</math> = <math>3</math>,
+<math>2 + b + c = 3</math>,
-<math>b</math> + <math>c</math> = <math>1</math>,
+<math>b + c = 1</math>,
-<math>b</math> = 0,1 ,
+<math>b = 0,1</math>,
-and respective values of <math>c</math> will be
+<math>c = 1,0</math>,
-<math>c</math> = 1,0 ,
-Which means <math>\boxed{\textbf\ 2}</math> solutions.
+<math>\boxed{\textbf\ 2}</math> solutions.
-Case <math>4</math>,
-<math>a</math> + <math>b</math> + <math>c</math> = <math>3</math>,
+Case <math>4:a = 3</math>,
-<math>a</math> = 3,
+<math>3 + b + c = 3</math>,
-<math>3</math> + <math>b</math> + <math>c</math> = <math>3</math>,
+<math>b + c = 0</math>,
-<math>b</math> + <math>c</math> = <math>0</math>,
+<math>b = 0</math>,
-<math>b</math> = 0 ,
+<math>c = 0</math>,
-and respective value of <math>c</math> will be
+<math>\boxed{\textbf\ 1}</math> solution.
-<math>c</math> = 0 ,
-Which means <math>\boxed{\textbf\ 1}</math> solution.
-Therefore there will be total 10 solutions. <math>\boxed{\textbf{(D)}\ 10}</math>.
-This is not a fast or an elegant solution but if this comes to your mind in the exam it will be beneficial.
+Therefore there will be a total  of 10 solutions. <math>\boxed{\textbf{(D)}\ 10}</math>.
 Solution by [[User:phoenixfire|phoenixfire]]

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