Difference between revisions of "2005 Alabama ARML TST Problems/Problem 12"
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==Solution== | ==Solution== | ||
=== Solution 1 === | === Solution 1 === | ||
| − | The [[generating function]] for <math>a, b, c,</math> and <math>d</math> is <math> | + | The [[generating function]] for <math>a, b, c,</math> and <math>d</math> is <math>x+x^2+x^3+\cdots</math>. |
<cmath>(x+x^2+x^3+\cdots)^4=x^4(1+x+x^2+x^3)^4= x^4+{4\choose 3}x^5+{5\choose 3}x^6+\cdots</cmath> | <cmath>(x+x^2+x^3+\cdots)^4=x^4(1+x+x^2+x^3)^4= x^4+{4\choose 3}x^5+{5\choose 3}x^6+\cdots</cmath> | ||
| − | The coefficient of <math>x^12</math> is <math>\binom{3+12-4}{3}= | + | The coefficient of <math>x^{12}</math> is <math>\binom{3+12-4}{3}=165</math>. |
=== Solution 2 === | === Solution 2 === | ||
| − | We consider a [[bijection]] to the following combinatorial argument. Suppose we have twelve balls and we would like to put them into four urns. There is a positive number of balls in each urn, so we put one ball into each urn initially. We are left with eight balls to go into four urns, which is equivalent to have three dividers. Thus there are <math>\frac{11!}{8!3!} = | + | We consider a [[bijection]] to the following combinatorial argument. Suppose we have twelve balls and we would like to put them into four urns. There is a positive number of balls in each urn, so we put one ball into each urn initially. We are left with eight balls to go into four urns, which is equivalent to have three dividers. Thus there are <math>\frac{11!}{8!3!} = 165</math> ordered pairs. |
==See also== | ==See also== | ||
{{ARML box|year=2005|state=Alabama|num-b=11|num-a=13}} | {{ARML box|year=2005|state=Alabama|num-b=11|num-a=13}} | ||
| + | <center>[[2006 Alabama ARML TST Problems/Problem 9|A similar problem]]</center> | ||
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
Latest revision as of 09:56, 18 June 2008
Problem
Find the number of ordered pairs of positive integers 
that satisfy the following equation:
.Solution
Solution 1
The generating function for 
and 
is 
.
![\[(x+x^2+x^3+\cdots)^4=x^4(1+x+x^2+x^3)^4= x^4+{4\choose 3}x^5+{5\choose 3}x^6+\cdots\]](http://latex.artofproblemsolving.com/a/f/6/af6b10a5b30af6b204cdd1721e90475bf7fec437.png)
The coefficient of 
is 
.
Solution 2
We consider a bijection to the following combinatorial argument. Suppose we have twelve balls and we would like to put them into four urns. There is a positive number of balls in each urn, so we put one ball into each urn initially. We are left with eight balls to go into four urns, which is equivalent to have three dividers. Thus there are 
ordered pairs.
See also
| 2005 Alabama ARML TST (Problems) | ||
| Preceded by: Problem 11 |
Followed by: Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
