Difference between revisions of "2011 AIME II Problems/Problem 6"
m (→Solution 2) |
m (→Solution 2) |
||
Line 15: | Line 15: | ||
==Solution 2== | ==Solution 2== | ||
− | Let us consider our quadruple <math>(a,b,c,d)</math> as the following image <math>xaxbcxxdxx</math>. The location of the letter <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> represents its value and <math>x</math> is a place holder. Clearly the quadruple is interesting if there are more place holders between <math>c</math> and <math>d</math> than there are between <math>a</math> and <math>b</math>. <math>0</math> holders between <math>a</math> and <math>b</math> means we consider <math>a</math> and <math>b</math> as one unit <math>ab</math> and <math>c</math> as <math>cx</math> yielding <math>\binom83 = 56</math> ways | + | Let us consider our quadruple <math>(a,b,c,d)</math> as the following image <math>xaxbcxxdxx</math>. The location of the letter <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> represents its value and <math>x</math> is a place holder. Clearly the quadruple is interesting if there are more place holders between <math>c</math> and <math>d</math> than there are between <math>a</math> and <math>b</math>. |
+ | |||
+ | <math>0</math> holders between <math>a</math> and <math>b</math> means we consider <math>a</math> and <math>b</math> as one unit <math>ab</math> and <math>c</math> as <math>cx</math> yielding <math>\binom83 = 56</math> ways; | ||
+ | |||
+ | <math>1</math> holder between <math>a</math> and <math>b</math> means we consider <math>a</math> and <math>b</math> as one unit <math>axb</math> and <math>c</math> as <math>cxx</math> yielding <math>\binom 63 = 20</math> ways; | ||
+ | |||
+ | <math>2</math> holders between <math>a</math> and <math>b</math> means we consider <math>a</math> and <math>b</math> as one unit <math>axxb</math> and <math>c</math> as <math>cxxx</math> yielding <math>\binom43 = 4</math> ways. | ||
+ | |||
+ | Since there cannot be <math>3</math> holders between <math>a</math> and <math>b</math> so our total is <math>56+20+4=\fbox{080}</math>. | ||
==Solution 3 (<i>Slightly</i> bashy)== | ==Solution 3 (<i>Slightly</i> bashy)== |
Revision as of 21:11, 5 March 2021
Contents
Problem 6
Define an ordered quadruple of integers as interesting if , and . How many interesting ordered quadruples are there?
Solution 1
Rearranging the inequality we get . Let , then is a partition of 11 into 5 positive integers or equivalently: 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 . 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 is the number of partitions where the second element is equal to the fourth, our answer is .
We find as a sum of 4 cases:
- two parts equal to zero, ways,
- two parts equal to one, ways,
- two parts equal to two, ways,
- two parts equal to three, way.
Therefore, and our answer is
Solution 2
Let us consider our quadruple as the following image . The location of the letter , , , represents its value and is a place holder. Clearly the quadruple is interesting if there are more place holders between and than there are between and .
holders between and means we consider and as one unit and as yielding ways;
holder between and means we consider and as one unit and as yielding ways;
holders between and means we consider and as one unit and as yielding ways.
Since there cannot be holders between and so our total is .
Solution 3 (Slightly bashy)
We first start out when the value of .
Doing casework, we discover that . We quickly find a pattern.
Now, doing this for the rest of the values of and , we see that the answer is simply:
Solution 4 (quick)
Notice that if , then , so there is a 1-to-1 correspondence between the number of ordered quadruples with and the number of ordered quadruples with .
Quick counting gives that the number of ordered quadruples with is 50. To count this, consider our numbers . Notice that if, for example, , that the average of and must both be . In this way, there is a symmetry for this case, centered at . If instead, say, , an odd number, then there is symmetry with about . Further, the number of cases for each of these centers of symmetry correspond to a triangular number. Eg centered at , there is case for each and so on, until centered at , there are possible cases. Adding these all, we have .
Thus the answer is
Solution 5
Think about a,b,c,and d as distinct objects, that we must place in 4 of 10 spaces. However, in only 1 of 24 of these combinations, will the placement of these objects satisfy the condition in the problem. So we know the total number of ordered quadruples is
Next, intuitively, the number of quadruples where is equal to the number of quadruples where . So we need to find the number of quadruples where the two quantities are equal. To do this, all we have to do is consider the cases when ranges from 3 to 9. It would seem natural that a range of 3 would produce 1 option, and a range of 4 would produce 2 options. However, since b and c cannot be equal, a range of 3 or 4 produces 1 option each, a range of 5 or 6 produces 2 options each, a range of 7 or 8 produces 3 options each, and a range of 9 will produce 4 options. In addition, a range of n has 10-n options for combinations of a and d. Multiplying the number of combinations of a and d by the corresponding number of options for b and c gives us 50 total quadruplets where .
So the answer will be
Solution 6
Let and and for positive integers In order to satisfy the other condition we need so we let Now the only other condition we need to satisfy so This condition can be transformed into for positive Now we use generating functions to finish. We find the generating function of the whole expression is and we are looking for the coefficient. This simplifies to finding the coefficient of Now this expression simplifies to The coefficient ends up to be
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
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.