2004 AMC 8 Problems/Problem 17
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
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 a1b2
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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