Difference between revisions of "2004 AMC 8 Problems/Problem 17"
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==Solution 2== | ==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. | 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 | + | Let the three friends be <math>a, b, c</math> repectively. |
| − | <math>a | + | <math>a + b + c = 3</math> |
The total being 3 and 2 plus signs, which implies | The total being 3 and 2 plus signs, which implies | ||
<math>\binom{3+2}{3} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}</math>. | <math>\binom{3+2}{3} = \binom{5}{3} = \boxed{\textbf{(D)}\ 10}</math>. | ||
Revision as of 01:22, 12 August 2020
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.
The total being 3 and 2 plus signs, which implies
.
Solution by phoenixfire
Solution 3
Like in solution 1 and solution 2, 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 casework.
Let the three friends be 
, 
, 
repectively.
Case 
+ + = ,
=0,
+ = ,
= 0,1,2,3 ,
and respective values of will be
= 3,2,1,0 ,
Which means 
solutions.
Case 
,
+ + = ,
=1,
+ + = ,
+ = ,
= 0,1,2 ,
and respective values of will be
= 2,1,0 ,
Which means 
solutions.
Case ,
+ + = ,
= 2,
+ + = ,
+ = ,
= 0,1 ,
and respective values of will be
= 1,0 ,
Which means 
solutions.
Case 
,
+ + = ,
= 3,
+ + = ,
+ = 
,
= 0 ,
and respective value of will be
= 0 ,
Which means 
solution.
Therefore there will be total 10 solutions. 
.
This is not a fast or an elegant solution but if this comes to your mind in the exam it will be beneficial.
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. 
