Difference between revisions of "2011 AIME II Problems/Problem 6"

(Problem 6)
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==Solution==
 
==Solution==
Rearranging the inequality we get <math>d-c > b-a</math>. Let <math>e = 11</math>, then <math>(a, b-a, c-b, d-c, e-d)</math> is a partition of 11 into 5 positive integers or equivalently:
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Rearranging the [[inequality]] we get <math>d-c > b-a</math>. Let <math>e = 11</math>, then <math>(a, b-a, c-b, d-c, e-d)</math> is a partition of 11 into 5 positive integers or equivalently:
<math>(a-1, b-a-1, c-b-1, d-c-1, e-d-1)</math> is a partition of 6 into 5 non-negative integer parts. Via a standard balls and urns argument, the number of ways to partition 6 into 5 non-negative parts is <math>\binom{6+4}4 = \binom{10}4 = 210</math>. The interesting quadruples correspond to partitions where the second number is less than the fourth. By symmetry there as many partitions where the fourth is less than the second. So, if <math>N</math> is the number of partitions where the second element is equal to the fourth, our answer is <math>(210-N)/2</math>.
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<math>(a-1, b-a-1, c-b-1, d-c-1, e-d-1)</math> is a [[partition]] of 6 into 5 non-negative integer parts. Via a standard balls and urns argument, the number of ways to partition 6 into 5 non-negative parts is <math>\binom{6+4}4 = \binom{10}4 = 210</math>. The interesting quadruples correspond to partitions where the second number is less than the fourth. By symmetry there as many partitions where the fourth is less than the second. So, if <math>N</math> is the number of partitions where the second element is equal to the fourth, our answer is <math>(210-N)/2</math>.
  
 
We find <math>N</math> as a sum of 4 cases:
 
We find <math>N</math> as a sum of 4 cases:
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* two parts equal to three, <math>\binom22 = 1</math> way.
 
* two parts equal to three, <math>\binom22 = 1</math> way.
 
Therefore, <math>N = 28 + 15 + 6 + 1 = 50</math> and our answer is <math>(210 - 50)/2 = \fbox{80.}</math>
 
Therefore, <math>N = 28 + 15 + 6 + 1 = 50</math> and our answer is <math>(210 - 50)/2 = \fbox{80.}</math>
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==See also==
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{{AIME box|year=2011|n=II|num-b=5|num-a=7}}
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 09:36, 23 August 2011

Problem 6

Define an ordered quadruple $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$, and $a+d>b+c$. How many interesting ordered quadruples are there?

Solution

Rearranging the inequality we get $d-c > b-a$. Let $e = 11$, then $(a, b-a, c-b, d-c, e-d)$ is a partition of 11 into 5 positive integers or equivalently: $(a-1, b-a-1, c-b-1, d-c-1, e-d-1)$ is a partition of 6 into 5 non-negative integer parts. Via a standard balls and urns argument, the number of ways to partition 6 into 5 non-negative parts is $\binom{6+4}4 = \binom{10}4 = 210$. The interesting quadruples correspond to partitions where the second number is less than the fourth. By symmetry there as many partitions where the fourth is less than the second. So, if $N$ is the number of partitions where the second element is equal to the fourth, our answer is $(210-N)/2$.

We find $N$ as a sum of 4 cases:

  • two parts equal to zero, $\binom82 = 28$ ways,
  • two parts equal to one, $\binom62 = 15$ ways,
  • two parts equal to two, $\binom42 = 6$ ways,
  • two parts equal to three, $\binom22 = 1$ way.

Therefore, $N = 28 + 15 + 6 + 1 = 50$ and our answer is $(210 - 50)/2 = \fbox{80.}$

See also

2011 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AIME Problems and Solutions
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