Difference between revisions of "2004 AMC 8 Problems/Problem 17"
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==Solution== | ==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 <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} = \binom{5}{ | + | For each person to have at least one pencil, assign one of the 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} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}</math>. |
==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}} | {{MAA Notice}} | ||
Revision as of 19:58, 15 October 2017
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
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 
.
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. 
