Difference between revisions of "2012 AMC 10B Problems/Problem 22"
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<math>\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{10}</math> | <math>\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{10}</math> | ||
| − | By the binomial theorem, this is <math>2^{9}</math> = <math>512</math>, or <math>(A)</math> | + | By the binomial theorem, this is <math>2^{9}</math> = <math>512</math>, or <math>(A)</math> |
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==Solution 2== | ==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 <math>(1,2,3,4,5...10)</math>. There are 9 arrows, so the answer is <math>2^{9}</math> = <math>512</math> | 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 <math>(1,2,3,4,5...10)</math>. There are 9 arrows, so the answer is <math>2^{9}</math> = <math>512</math> | ||
Revision as of 20:15, 3 January 2014
Contents
Problem 22
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 one goes, and the numbers ascend from the first number, 2, with the exception of the 1.
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.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
