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 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 by phoenixfire
Minor Edits by Yuvag : "... is ." to "... is ."
All credit still goes to phoenixfire.
Solution 2
Like in solution 1, for each person to have at least one pencil, assign one of the pencils to each of the three friends so that you have left. In partitioning the remaining pencils into distinct groups, use number of non-negative integral solutions. Let the three friends be respectively.
The total being 3 and 2 plus signs, which implies .
Solution by phoenixfire
Minor Edits by Yuvag : "." to ".
All credit still goes to phoenixfire.
Solution 3
For each of the 3 People to have at least one pencil when distributing 6 pencil amongst them, we can use another formula from the Ball-and-urn counting technique, shown below:
for n = number of items, and s = slots:
Now we can plug in our values,
number of items = 6, and slots = 3:
.
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 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 solutions. . Solution by phoenixfire
Video Solution
https://youtu.be/FUnwTLP7gr0 Soo, DRMS, NM
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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