2004 AMC 8 Problems/Problem 17: Difference between revisions
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<math>\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12</math> | <math>\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12</math> | ||
==Solution== | ==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 <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>. | 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>. | ||
==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 <math>3</math> left. In partitioning the remaining <math>3</math> pencils into <math>3</math> distinct groups, use number of non-negetive integral soutions. | |||
Let the three friends be <math>a</math>, <math>b</math>, <math>c</math> repectively. | |||
<math>a</math> + <math>b</math> + <math>c</math> = 3 | |||
The total being 3 and 2 plus signs, which implies | |||
<math>\binom{3+2}{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 10:53, 15 November 2019
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.
+ + = 3
The total being 3 and 2 plus signs, which implies
.
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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