# 2006 AIME II Problem 4-COMBINATORICS

PROBLEM:

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

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.

I did understand the problem and and used this sort of reasoning:

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 $${11 \choose 5}=462$$ such ordered 12-tuples.

I looked up the internet and also saw this solution:

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 $$a_6, only half of the choices will work, so the answer is $$\frac{\binom{12}{6}}{2}=462$$ 12-tuples

however,i wasn't able to understand this approach..Please explain the second approach.

Thank you in advance (P.S i am an ardent believer in the fact that knowing multiple ways to attack a certain problem sharpens problem solving skills)

The chosen six are $$a_1,...,a_6$$, and $$a_6$$ is the lowest of those. The other six are $$a_7,...,a_{12}$$, and $$a_7$$ is the lowest of those. The only thing to check is that $$a_6\lt a_7$$. By symmetry between the two groups of six, they each have the very lowest number equally, so $$a_6\lt a_7$$ half the time.