2006 AIME II Problems/Problem 4

Revision as of 15:30, 29 February 2020 by I Love Math (talk | contribs) (box the solution)


Let $(a_1,a_2,a_3,\ldots,a_{12})$ be a permutation of $(1,2,3,\ldots,12)$ for which

$a_1>a_2>a_3>a_4>a_5>a_6 \mathrm{\  and \ } a_6<a_7<a_8<a_9<a_{10}<a_{11}<a_{12}.$

An example of such a permutation is $(6,5,4,3,2,1,7,8,9,10,11,12).$ Find the number of such permutations.


Clearly, $a_6=1$. Now, consider selecting $5$ of the remaining $11$ values. Sort these values in descending order, and sort the other $6$ values in ascending order. Now, let the $5$ selected values be $a_1$ through $a_5$, and let the remaining $6$ be $a_7$ through ${a_{12}}$. It is now clear that there is a bijection between the number of ways to select $5$ values from $11$ and ordered 12-tuples $(a_1,\ldots,a_{12})$. Thus, there will be ${11 \choose 5}=\boxed{462}$ such ordered 12-tuples.

Solution 2

There are $\binom{12}{6}$ ways to choose 6 numbers from $(1,2,3,\ldots,12)$, and then there will only be one way to order them. And since that $a_6<a_7$, only half of the choices will work, so the answer is $\frac{\binom{12}{6}}{2}=462$ 12-tuples - mathleticguyyy

See also

2006 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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