Difference between revisions of "2005 Alabama ARML TST Problems/Problem 12"

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<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}=455</math>.
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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!} = 455</math> ordered pairs.
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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}}
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<center>[[2006 Alabama ARML TST Problems/Problem 9|A similar problem]]</center>
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]

Latest revision as of 08:56, 18 June 2008

Problem

Find the number of ordered pairs of positive integers $(a,b,c,d)$ that satisfy the following equation:

$a+b+c+d=12$.

Solution

Solution 1

The generating function for $a, b, c,$ and $d$ is $x+x^2+x^3+\cdots$.

\[(x+x^2+x^3+\cdots)^4=x^4(1+x+x^2+x^3)^4= x^4+{4\choose 3}x^5+{5\choose 3}x^6+\cdots\]

The coefficient of $x^{12}$ is $\binom{3+12-4}{3}=165$.

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 $\frac{11!}{8!3!} = 165$ 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
A similar problem