2004 AMC 8 Problems/Problem 17

Revision as of 12:07, 15 November 2019 by Phoenixfire (talk | contribs) (Solution 3)

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.

Case $1$ $a$ + $b$ + $c$ = $3$, $a$ =0, $b$ + $c$ = $3$, $b$ = 0,1,2,3 , and respective values of $c$ will be $c$ = 3,2,1,0 , Which means 4 solutions.

Case $2$, $a$ + $b$ + $c$ = $3$, $a$ =1, $1$ + $b$ + $c$ = $3$, $b$ + $c$ = $2$, $b$ = 0,1,2 , and respective values of $c$ will be $c$ = 2,1,0 , Which means 3 solutions.

Case $3$ , $a$ + $b$ + $c$ = $3$, $a$= 2, $2$ + $b$ + $c$ = $3$, $b$ + $c$ = $1$, $b$ = 0,1 , and respective values of $c$ will be $c$ = 1,0 , Which means 2 solutions.

Case $4$, $a$ + $b$ + $c$ = $3$, $a$ = 3, $3$ + $b$ + $c$ = $3$, $b$ + $c$ = $0$, $b$ = 0 , and respective value of $c$ will be $c$ = 0 , Which means 1 solution.

Therefore there will be total 10 solutions. \boxed{\textbf{(D)}\ 10}.

This is not a fast or an elegant solution but if this comes to your mind in the exam it will be beneficial.

Solution by $phoenixfire$

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
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

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