2012 AMC 10B Problems/Problem 22
Contents
Problem
Let (,
, ...
) be a list of the first 10 positive integers such that for each
either
or
or both appear somewhere before
in the list. How many such lists are there?
Solution 1
If we have 1 as the first number, then the only possible list is .
If we have 2 as the first number, then we have 9 ways to choose where the goes, and the numbers ascend from the first number,
, with the exception of the
.
For example,
, or
. There are
ways to do so.
If we use 3 as the first number, we need to choose 2 spaces to be 2 and 1, respectively. There are ways to do this.
In the same way, the total number of lists is:
By the binomial theorem, this is =
, or
Solution 2
Arrange the spaces and put arrows pointing either up or down between them. Then for each arrangement of arrows there is one and only one list that corresponds to up. For example, all arrows pointing up is . There are 9 arrows, so the answer is
=
NOTE: Solution cited from: http://www.artofproblemsolving.com/Videos/external.php?video_id=269.
Solution 3
Notice that the answer to the problem is solely based on the length of the lists, i.e. 10. We can replace 10 with smaller values, such as 2 and 3, and try to find a pattern. If we replace it with 2, we can easily see that there are two possible lists, and
. If we replace it with 3, there are four lists,
and
. Since 2 and 4 are both powers of 2, it is likely that the number of lists is
, where
is the length of the lists.
Solution 4 (Recursion)
If , the sequence must be
. If
, then
, and the sequence is
. If
, then the possible sequences are
In general, for an
-length sequence, if
, then
through
can be filled in
ways with
through
, and
through
must be sorted in decreasing order with the remaining numbers (
through
), in one way. Thus
, where
.
We can see (or prove by induction) that . Hence,
.
- ColtsFan10
Solution 5
Assume the same conditions to be held and let's look at several smaller cases to find a pattern. If we are only arranging there are trivially only
ways. Now let us look at arranging
. You can arrange this in
ways. Looking at
you can arrange this in
ways. The pattern becomes evident now. If there are
numbers there are
ways. Hence our answer would be
ways which is
.
-srisainandan6
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012amc10b/269
~dolphin7
Video Solution by TheBeautyofMath
~IceMatrix
See Also
2012 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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