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PROBLEM:

Let $(a_1,a_2,a_3,\ldots,a_{12})$ be a permutation of $(1,2,3,\ldots,12)$ for which $a_1>a_2>a_3>a_4>a_5>a_6 \mathrm{\ and \ } a_6<a_7<a_8<a_9<a_{10}<a_{11}<a_{12}.$

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.

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 $a_6<a_7$, 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)

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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.

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