国际象棋社团合影

先安排中间位置。中间共有 $3$ 名成员，他们在这 3 个位置上的排列方式共有 $3! = 6$ 种。

再看某一侧。两侧总共要站 8 个人，因此每一侧各有 4 人。11 人中已有 3 人被安排在中间，剩下 8 人可供选择。对于第一侧，我们从这 8 人中任取 4 人即可。这给出 $\binom{8}{4} = 70$ 种选人方式。选好后，这 4 个人有 2 种排法：升序或降序，因此第一侧共有 $2\cdot70=140$ 种安排。

另一侧直接由剩下的 4 人组成，不需要再考虑选人方式。这 4 人同样有 2 种排列方式。

因此总排列数为：

$$6\cdot 2\cdot\binom{8}{4}\cdot 2=\boxed{1680}.$$

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Original Explanation

Let's start with arranging the center. There are $3$ members in the center and they can be placed any way among these three spots resulting in a total of $3!$ or $6$ different centers.

Now let's look at a side. Eight total people go on the sides but on one particular side we will have four people. Out of the $11$ total members, $3$ have already been placed in the center so we have $8$ remaining to select from. Of those $8$, we can select any $4$ as there will exist a way to sequentially order any number of unique ratings. This gives us $8\choose4$ or $70$ ways to pick members for the first side. Once these members have been selected, there are $2$ ways of arranging them: ascending or descending order. Giving us a total of $2\cdot70$ or $140$ orderings within our first side.

We put all of our remaining members on the other side so we don't have to worry about how many ways to select them. There will also be $2$ total ways of arranging these last $4$ members.

Hence the total number of arrangements is:

$$6 \cdot 2{8\choose4}\cdot2 = \boxed{1680}$$