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

(Solution)
(Solution 2)
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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>
 
==Solution 2==
 
==Solution 2==
Let us consider our quadruple (a,b,c,d) as the following image  xaxbcxxdxx. The location of the letter a,b,c,d represents its value and x is a place holder. Clearly the quadruple is interesting if there are more place holders between c and d then there are between a and b. 0 holders between a and b means we consider a and b as one unit ab and c as cx yielding <math>\binom83 = 56</math> ways, 1 holder between a and b means we consider a and b as one unit axb and c as cxx yielding <math>\binom 63 = 20</math> ways, 2 holders between a and b means we consider a and b as one unit axxb and c as cxxx yielding <math>\binom43 = 4</math> ways and there cannot be 3 holders between a and b so our total is 56+20+4=<math>\fbox{80.}</math>.
+
Let us consider our quadruple (a,b,c,d) as the following image  xaxbcxxdxx. The location of the letter a,b,c,d represents its value and x is a place holder. Clearly the quadruple is interesting if there are more place holders between c and d than there are between a and b. 0 holders between a and b means we consider a and b as one unit ab and c as cx yielding <math>\binom83 = 56</math> ways, 1 holder between a and b means we consider a and b as one unit axb and c as cxx yielding <math>\binom 63 = 20</math> ways, 2 holders between a and b means we consider a and b as one unit axxb and c as cxxx yielding <math>\binom43 = 4</math> ways and there cannot be 3 holders between a and b so our total is 56+20+4=<math>\fbox{80.}</math>.
  
 
==See also==
 
==See also==

Revision as of 21:06, 20 April 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 1

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 $(210 - 50)/2 = \fbox{80.}$

Solution 2

Let us consider our quadruple (a,b,c,d) as the following image xaxbcxxdxx. The location of the letter a,b,c,d represents its value and x is a place holder. Clearly the quadruple is interesting if there are more place holders between c and d than there are between a and b. 0 holders between a and b means we consider a and b as one unit ab and c as cx yielding $\binom83 = 56$ ways, 1 holder between a and b means we consider a and b as one unit axb and c as cxx yielding $\binom 63 = 20$ ways, 2 holders between a and b means we consider a and b as one unit axxb and c as cxxx yielding $\binom43 = 4$ ways and there cannot be 3 holders between a and b so our total is 56+20+4=$\fbox{80.}$.

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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