2011 AIME II Problems/Problem 6

Revision as of 18:44, 14 May 2019 by Inconsistent (talk | contribs) (Solution 3 (Slightly bashy): Spacing out the lines to prevent overflow off the page.)

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 stars and bars 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 are 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 $(210 - 50)/2 = \fbox{080}$

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=$\fbox{080}$.

Solution 3 (Slightly bashy)

We first start out when the value of $a=1$.

Doing casework, we discover that $d=5,6,7,8,9,10$. We quickly find a pattern.

Now, doing this for the rest of the values of $a$ and $d$, we see that the answer is simply:

$(1)+(2)+(1+3)+(2+4)+(1+3+5)+(2+4+6)+(1)+(2)+(1+3)+(2+4)$ $+(1+3+5)+(1)+(2)+(1+3)+(2+4)+(1)+(2)+(1+3)+(1)+(2)+(1)=\boxed{080}$

Solution 4 (quick)

Notice that if $a+d>b+c$, then $(11-a)+(11-d)<(11-b)+(11-c)$, so there is a 1-to-1 correspondence between the number of ordered quadruples with $a+d>b+c$ and the number of ordered quadruples with $a+d<b+c$.

Quick counting gives that the number of ordered quadruples with $a+d=b+c$ is 50.

Thus the answer is $\frac{\binom{10}{4}-50}{2} = \boxed{080}.$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS