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

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(Problem 6)
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==Problem 6==
 
==Problem 6==
  
Define an ordered quadruple <math>(a, b, c, d)</math> as interesting if <math>1 \le a<b<c<d \le 10</math>, and <math>a+d>b+c</math>. How many interesting ordered quadruples are there?
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Define an ordered quadruple of integers <math>(a, b, c, d)</math> as interesting if <math>1 \le a<b<c<d \le 10</math>, and <math>a+d>b+c</math>. How many interesting ordered quadruples are there?
  
 
==Solution==
 
==Solution==

Revision as of 20:40, 9 January 2012

Problem 6

Define an ordered quadruple of integers $(a, b, c, d)$ 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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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