Difference between revisions of "2004 AMC 8 Problems/Problem 17"
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==Solution 1== | ==Solution 1== | ||
For each person to have at least one pencil, assign one pencil to each of the three friends so that you have <math>3</math> left. In partitioning the remaining <math>3</math> pencils into <math>3</math> distinct groups, use [[Ball-and-urn]] to find the number of possibilities is <math>\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>. | For each person to have at least one pencil, assign one pencil to each of the three friends so that you have <math>3</math> left. In partitioning the remaining <math>3</math> pencils into <math>3</math> distinct groups, use [[Ball-and-urn]] to find the number of possibilities is <math>\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>. | ||
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+ | Solution by [[User:phoenixfire|phoenixfire]] | ||
+ | Minor Edits by [[User:Yuvag|Yuvag]] : "... is <math>\binom{3+3-1}{3} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}</math>." to "... is <math>\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{\textbf{(D)}\ 10}</math>." | ||
+ | |||
+ | All credit still goes to phoenixfire. | ||
==Solution 2== | ==Solution 2== |
Revision as of 12:07, 4 January 2023
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 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
For each person to have at least one pencil, we can use another formula from the Ball-and-urn counting technique:
for n = number for item, and s = slots:
.
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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.