Difference between revisions of "2012 AMC 12B Problems/Problem 18"
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<cmath>\binom{9}{0} + \binom{9}{1} + \cdots + \binom{9}{9} = 2^9=512 ...... \framebox{B}</cmath> | <cmath>\binom{9}{0} + \binom{9}{1} + \cdots + \binom{9}{9} = 2^9=512 ...... \framebox{B}</cmath> | ||
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+ | == See Also == | ||
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+ | {{AMC12 box|year=2012|ab=B|num-b=17|num-a=19}} |
Revision as of 22:21, 12 January 2013
Problem 18
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
Let . Assume that . If , the first number appear after that is greater than must be , otherwise if it is any number larger than , there will be neither nor appearing before . Similarly, one can conclude that if , the first number appear after that is larger than must be , and so forth.
On the other hand, if , the first number appear after that is less than must be , and then , and so forth.
To count the number of possibilities when is given, we set up spots after , and assign of them to the numbers less than and the rest to the numbers greater than . The number of ways in doing so is choose .
Therefore, when summing up the cases from to , we get
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |