Difference between revisions of "2009 AIME II Problems/Problem 6"
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Revision as of 22:36, 4 July 2013
Problem
Let 
be the number of five-element subsets that can be chosen from the set of the first 
natural numbers so that at least two of the five numbers are consecutive. Find the remainder when is divided by 
.
Solution
We can use complementary counting. We can choose a five-element subset in 
ways. We will now count those where no two numbers are consecutive. We will show a bijection between this set, and the set of 10-element strings that contain 5 
s and 5 
s, thereby showing that there are 
such sets.
Given a five-element subset 
of 
in which no two numbers are consecutive, we can start by writing down a string of length 14, in which the 
-th character is if 
and otherwise. Now we got a string with 5 s and 9 s. As no two numbers were consecutive, we know that in our string no two s are consecutive. We can now remove exactly one from between each pair of s to get a string with 5 s and 5 s. And clearly this is a bijection, as from each string with 5 s and 5 s we can reconstruct one original set by reversing the construction.
Hence we have 
, and the answer is 
.
See Also
| 2009 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. 
