2004 AMC 8 Problems/Problem 17
Problem
Three friends have a total of identical pencils, and each one has at least one pencil. In how many ways can this happen?
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 left. In partitioning the remaining pencils into distinct groups, use Ball-and-urn to find the number of possibilities is .
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 left. In partitioning the remaining pencils into distinct groups, use number of non-negetive integral soutions. Let the three friends be repectively.
The total being 3 and 2 plus signs, which implies .
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 left. In partitioning the remaining pencils into distinct groups use casework. Let the three friends be , , repectively.
,
Case ,
,
,
,
solutions.
Case ,
,
,
,
,
solutions.
Case ,
,
,
,
,
solutions.
Case ,
,
,
,
,
solution.
Therefore there will be a total of 10 solutions. . Solution by phoenixfire
See Also
2004 AMC 8 (Problems • Answer Key • Resources) | ||
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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