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- | + | 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-turn]] to find the number of possibilities is <math>_{3+3-1} C _{3-1} = _10 C _2 = \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 12:17, 28 October 2016
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-turn 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 |
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