Art of Problem Solving
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2004 AMC 8 Problems/Problem 17: Difference between revisions

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==See Also==
==See Also==
{{AMC8 box|year=2004|num-b=16|num-a=18}}
{{AMC8 box|year=2004|num-b=16|num-a=18}}
{{MAA Notice}}

Revision as of 00:00, 5 July 2013

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

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 stars and bars to find the number of possibilities is $_{3+3-1} C _{3-1} = _10 C _2 = \boxed{\textbf{(D)}\ 10}$.

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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